ELEMENTS    OF 

ENGINEERING 
THERMODYNAMICS 


BY 
JAMES   A.    MOYER 

Director  of  the  Massachusetts  Department  of  University  F.  ^tension 

formerly  Professor  of  Mechanical  Engineering  in  the 

Pennsylvania  State  College 

JAMES   P.  CALDERWOOD 

Professor  of  Mechanical  Engineering  in  the  Kansas  State 
Agricultural  College 

ANDREY  A.  POTTER 

Dean  of  Engineering  at  Purdue  University,  formerly 

Dean  of  Engineering  at  the  Kansas  State 

Agricultural  College 


NEW  YORK 

JOHN  WILEY  &  SONS,  INC. 

LONDON:    CHAPMAN  &  HALL,  LIMITED 
IQ2O 


f 


i^ngineeriag 
Library 


COPYRIGHT,  1920,  BY 
JAMES  A.  MOYER,  JAMES  P.  CALDERWOOD 

AND 

ANDREY  A.  POTTER 


Manufactured  in  the  U.S.  A. 


PREFACE 

This  treatise  is  an  extension  of  a  briefer  work  entitled  "  Engi- 
neering Thermodynamics"  by  James  A.  Moyer  and  James  P. 
Calderwood.  The  additions  and  changes  are  made  to  suit  the 
needs  of  those  who  had  successful  experience  in  using  the  original 
publication  and  found  it  desirable  to  add  supplementary  material 
to  make  it  sufficiently  inclusive  for  special  institutional  require- 
ments. A  great  deal  of  the  new  material  is  supplied  by  A,  A. 
Potter. 

This  book  is  intended  to  bring  out  the  fundamental  principles 
of  Engineering  Thermodynamics,  and  is  particularly  intended 
for  use  in  technical  colleges  where  it  is  possible  to  give  special 
courses  on  the  subjects  of  steam  turbines,  internal  combustion 
engines,  refrigeration  and  other  applications  of  thermody- 
namics. 

The  new  material  includes  the  theory  of  the  hot  air  engine 
and  internal  combustion  engine  cycles.  The  appendix  includes 
logarithmic  tables,  the  properties  of  gases,  the  properties  of 
steam,  ammonia,  sulphur  dioxide  and  carbon  dioxide. 

Every  engineering  student  should  become  familiar  with' stand- 
ard works  on  the  subject  of  thermodynamics.  This  book  should 
consequently  be  supplemented  by  references  to  standard  works  on 
this  subject.  A  representative  list  of  such  reference  books  is 
given  in  a  table  of  the  appendix. 

The  authors  are  particularly  indebted  to  the  following  pro- 
fessors and  instructors  in  Mechanical  Engineering  Departments 
who  gave  valuable  suggestions  and  criticisms:  Edwin  A.  Fessen- 
den,  Pennsylvania  State  College;  J.  E.  Emswiler,  University  of 
Michigan;  G.  L.  Chris tensen,  Michigan  College  of  Mines;  Roy 

iii 


^47194 


iv  PREFACE 

B.  Fehr,  formerly  of  Pennsylvania  State  College;  H.  L.  Seward, 
Yale  University;  J.  R.  Wharton,  University  of  Missouri;  and 
J.  J.  Wilmore,  Alabama  Polytechnic  Institute. 

J.  A.   MOYER, 

J.  P.   CALDERWOOD, 

A.  A.  POTTER. 


CONTENTS 


PAGES 
CHAPTER  I.  —  THERMODYNAMIC  PRINCIPLES  AND  DEFINITIONS 1-13 

Definition,  Scope  and  Object  of  Thermodynamics;  Heat;  Temperature; 
Thermometers;  Absolute  Zero;  Units  of  Heat;  Specific  Heat;  Pressure; 
Volume;  Work;  Power;  Indicated  Horse-power;  Brake  Horse-power; 
Mechanical  Equivalent  of  Heat;  First  Law  of  Thermodynamics;  Second 
Law  of  Thermodynamics;  Effects  of  Heat  Application;  The  Heat  Engine; 
Thermal  Efficiency  of  a  Heat  Engine;  Problems. 

CHAPTER  II.  —  PROPERTIES  or  PERFECT  GASES 14-25 

Perfect  Gas  Defined;  Relation  between  Pressure,  Volume,  and  Temper- 
ature of  a  Perfect  Gas;  Boyle's  Law;  Charles'  Law;  Combination  of 
Boyle's  and  Charles'  Laws;  The  Law  of  Perfect  Gases;  Heat  and  its 
Effect  upon  a  Gas;  External  Work;  Internal  Energy;  Joule's  Law; 
Relation  of  Specific  Heats  and  the  Gas  Constant;  Ratio  of  the  Two 
Specific  Heats;  Values  of  the  Specific  Heats;  Problems. 

CHAPTER  III.  —  EXPANSION  AND  COMPRESSION  OF  GASES 26-39 

Expansion  at  Constant  Pressure;  Expansion  at  Constant  Volume; 
Expansion  at  Constant  Temperature,  or  Isothermal;  Adiabatic  Expan- 
sion and  Compression;  Change  of  Internal  Energy  during  Adiabatic 
Processes;  Relation  between  Volume,  Pressure  and  Temperature  in 
Adiabatic  Expansion  of  a  Perfect  Gas;  Problems. 

CHAPTER  IV.  —  CYCLES  OF  HEAT  ENGINES  USING  GAS 40-61 

(1)  The  Carnot  Cycle;  Reversible  Cycles. 

(2)  Hot  Air  Engine  Cycle.    The  Stirling  Engine;  The  Ericsson  Engine. 

(3)  Internal  Combustion  Engine  Cycles.    The  Lenoir  Engine;  The  Otto 

Cycle;  The  Brayton  Cycle;  the  Diesel  Cycle. 
Problems. 

CHAPTER  V.  —  PROPERTIES  OF  VAPORS 62-90 

Saturated  and  Superheated  Vapors;  Theory  of  Vaporization;  Vapor 
Tables;  Relation  between  Temperature,  Pressure  and  Volume  of  Satu- 
rated Steam;  Heat  in  the  Liquid  (Water);  Latent  Heat  of  Evaporation; 
External  Work  of  Evaporation;  Total  Heat  of  Steam;  Internal  Energy 
of  Evaporation  and  of  Steam;  Steam  formed  at  Constant  Volume;  Wet 
Steam;  Superheated  Steam;  Drying  of  Steam  by  Throttling  or  Wire- 


vi  CONTENTS 

PAGES 
drawing;    Determination    of   the    Moisture  in   Steam;    Throttling   or 

Superheating  Calorimeter;  Barrus  Throttling  Calorimeter;  Separating 
Calorimeters;  Condensing  or  Barrel  Calorimeter;  Equivalent  Evapo- 
ration and  Factor  of  Evaporation;  Vapors  as  Refrigerating  Media; 
Problems. 

CHAPTER  VI.  —  ENTROPY 91-103 

Entropy  Changes  during  Constant  Pressure  Expansions  of  Gases; 
Entropy  Changes  of  Gases  at  Constant  Volume;  Entropy  Changes 
during  Isothermal  Processes  of  a  Gas;  Entropy  Changes  during  Reversible 
Adiabatic  Processes  of  Gases;  Entropy  Changes  during  a  Carnot  Cycle; 
Temperature- Entropy  Diagrams  for  Steam;  Calculation  of  Entropy  for 
Steam;  Total  Entropy  of  Steam;  The  Mollier  Chart;  Temperature- 
Entropy  Diagram  for  the  Steam  Power  Plant;  Problems. 

CHAPTER  VII.  —  EXPANSION  AND  COMPRESSION  OF  VAPORS 104-116 

Expansion  of  Wet  Steam  at  Constant  Volume;  Expansion  of  Super- 
heated Steam  at  Constant  Volume;  Expansion  at  Constant  Pressure; 
Isothermal  Lines  of  Steam;  Adiabatic  Lines  for  Steam;  Quality  of 
Steam  during  Adiabatic  Expansion;  Graphical  Determination  of  Quality 
of  Steam  by  Throttling  Calorimeter  and  Total  Heat-Entropy  Diagram; 
Poly  tropic  Expansion  (n  =  i);  Problems. 

CHAPTER  VIII.  —  CYCLES  OF  HEAT  ENGINES  USING  VAPORS 117-149 

Carnot  Cycle;  Rankine  Cycle;  Practical  or  Actual  Steam  Engine  Cycle; 
Efficiency  of  an  Engine  Using  Steam  Without  Expansion;  Adiabatic 
Expansion  and  Available  Energy;  Available  Energy  of  Wet  Steam; 
Available  Energy  of  Superheated  Steam;  Application  of  Temperature- 
Entropy  Diagram  to  Analysis  of  Steam  Engine;  Combined  Indicator 
Card  of  Compound  Engine;  Hirn's  Analysis;  Clayton 's  Analysis;  Prob- 
lems. 

CHAPTER  IX.  —FLOW  OF  FLUIDS 150-177 

Flow  Through  a  Nozzle  or  Orifice;  Weight  per  Cubic  Foot;  Maximum 
Discharge;  Shape  of  Nozzle;  Flow  of  Air  Through  an  Orifice;  Receiver 
Method  of  Measuring  Air;  Flow  of  Vapors;  Velocity  of  Flow  as  Affected 
by  Radiation;  Friction  Loss  in  a  Nozzle;  Impulse  Nozzles;  Turbine 
Losses;  Reaction  Nozzles;  Coefficient  of  Flow;  Injectors;  Weight  of 
Feed  Water  Supplied  by  an  Injector  per  Pound  of  Steam;  Thermal  Effi- 
ciency of  an  Injector;  Mechanical  Efficiency  of  an  Injector;  Orifice 
Measurements  of  the  Flow  of  Steam;  Flow  of  Steam  Through  Nozzles; 
Flow  of  Steam  when  the  Final  Pressure  is  more  than  0.58  of  the  Initial 
Pressure;  Length  of  Nozzles;  Efficiency  of  Nozzles;  Under-  and  Over- 
Expansion;  Non-expanding  Nozzles;  Materials  for  Nozzles;  Problems. 


CONTENTS  Vli 

PAGE 
CHAPTER  X.  —  APPLICATIONS  OF  THERMODYNAMICS  TO  COMPRESSED  AIR 

AND  REFRIGERATING  MACHINERY 178-196 

Compressed  Air;  Air  Compressors;  Work  of  Compression;  Effect  of 
Clearance  upon  Volumetric  Efficiency;  Two  Stage  Compression;  Re- 
frigerating Machines  or  Heat  Pumps;  Unit  of  Refrigeration;  Systems 
of  Mechanical  Refrigeration;  The  Air  System  of  Refrigeration;  The 
Vapor  Compression  System  of  Refrigeration;  The  Vapor  Absorption 
System  of  Refrigeration;  Coefficient  of  Performance  of  Refrigerating 
Machines;  Problems. 


SYMBOLS 


A  =  area  in  square  feet,  also  used  to  represent  the  reciprocal  of  the  mechanical 

equivalent  of  heat,  7^. 

B.t.u.  =  British  thermal  units  (=  778  ft.-lbs.). 

Cp  =  specific  heat  at  constant  pressure  in  B.t.u.  per  pound  per  degree. 
Cn  —  specific  heat  at  constant  volume  in  B.t.u.  per  pound  per  degree. 
C  =  a  general  constant  in  equations  of  perfect  gases. 

E  =  external  work  in  B.t.u.  per  pound;   also  sometimes  used  to  express  effi- 
ciency, usually  as  a  decimal. 
Ea  =  available  energy  hi  B.t.u.  per  pound. 
F  =  force  in  pounds. 
H  =  heat  per  pound  in  B.t.u.* 
#sup.  =  total  heat  of  superheated  steam,  B.t.u.  per  pound. 

IH  =  total  internal  energy  of  steam  (above  32°  F.)  in  B.t.u.  per  pound,  some- 
times designated  by  E  and  i". 
IL  —  internal  energy  of  evaporation  of  steam  in  B.t.u.  per  pound,  sometimes 

designated  by  p. 

J  =  mechanical  equivalent  of  heat  =  778  (use  becoming  obsolete). 
K  =  specific  heat  in  foot-pound  units. 
L  =  latent  heat  of  evaporation  in  B.t.u.  per  pound. 
P  =  pressure  in  general  or  pressure  in  pounds  per  square  foot. 
Q  =  quantity  of  heat  in  B.t.u. 
R  =  thermodynamic  constant  for  gases;  for  air  it  is  53.3  (in  foot-pound  units 

per  pound.) 

T  =  absolute  temperature,  in  Fahr.  degrees  =  460  +  L 

V  =  volume  in  cubic  feet,  also  specific  volume  and  velocity  in  feet  per  second. 
W  =  work  done  in  foot-pounds. 
a  =  area  in  square  inches. 
c  =  constant  of  integration. 
d  =  distance  in  feet. 

e  =  subscript  to  represent  base  of  natural  logarithms. 
g  =  acceleration  due  to  gravity  =  32.2  feet  per  second  per  second. 
h  =  heat  of  the  liquid  per  pound  in  B.t.u.  (above  32°  F.)  also  designated 

by  q  and  i'. 

i'  =  heat  of  liquid  hi  B.t.u.  per  pound  (above  32°  F.). 
i"  =  total  internal  energy  of  steam. 
k  =  a  constant, 
log  =  logarithm  to  base  10. 
\oge  =  logarithm  to  natural  base  e  (Naperian). 

*  In  steam  tables  it  is  total  heat  above  32°  F. 
ix 


SYMBOLS 

n  =  general  exponent  for  V  (volume)  in  equations  of  perfect  gases,  also  some- 
times used  for  entropy  of  the  liquid  in  B.t.u.  per  degree  of  absolute  tem- 
perature. 

p  =  pressure  in  pounds  per  square  inch. 

q  =  sometimes  used  for  heat  of  the  liquid  in  B.t.u.  per  pound  (above  32°  F.). 

r  ==  ratio  of  expansion  also  sometimes  used  for  latent  heat  of  evaporation  in 
B.t.u.  per  pound. 

s'  =  entropy  of  the  liquid. 
s"  =  total  entropy  of  vapors. 

/  =  temperature  in  ordinary  Fahr.  degrees. 

u  =  difference  between  the  specific  volume  of  a  vapor  and  that  of  the  liquid 
from  which  it  is  formed. 

v  =  specific  volume,  in  cubic  feet  per  pound  (in  some  steam  tables). 

w  =  weight  per  cubic  foot  =  density,  also  used  to  designate  weight  of  a  sub- 
stance in  pounds. 

x  =  quality  of  steam  expressed  as  a  decimal. 

ft 

y  =  ratio  of  specific  heats  -£. 
Co 

0  =  total  entropy  -^,  sometimes  designated  by  S". 

6  =  entropy  of  the  liquid  in  B.t.u.  per  pound  per  degree  of  absolute  tem- 
perature. 

p  =  internal  energy  of  evaporation,  B.t.u.  per  pound. 
<r  =  volume  of  water  in  a  cubic  foot  of  saturated  steam. 
"  =  inches. 


ELEMENTS    OF 

Engineering  Thermodynamics 


CHAPTER   I 
THERMODYNAMIC  PRINCIPLES  AND  DEFINITIONS 

Thermodynamics.  Thermodynamics  deals  with  the  relation 
between  heat  and  mechanical  work.  The  object  of  the  study  of 
thermodynamics  is  to  consider  factors  which  influence  the 
efficiency  of  heat  power  machinery. 

Thermodynamics  makes  it  possible  to  predict  the  perform- 
ance of  steam  engines  and  steam  turbines  when  operating  under 
conditions  of  increased  pressure,  of  higher  vacuums  and  expan- 
sions, of  higher  superheats,  of  jacketing  cylinders,  of  using  the 
working  medium  in  several  cylinders  one  after  the  other,  —  that 
is,  compounding  instead  of  expanding  the  steam  only  in  a  single 
cylinder,  —  of  inserting  receivers  between  the  cylinders  and  of 
reheating  the  working  medium  as  it  passes  from  one  cylinder  to 
the  next. 

By  means  of  calculations  based  upon  the  study  of  thermo- 
dynamics it  is  possible  to  determine  the  effect  of  increasing  the 
compression  pressures  of  a  gas  engine  mixture  before  it  is 
ignited,  the  result  of  incomplete  cooling  upon  the  efficiency  of 
an  air  compressor,  and  similar  problems. 

Another  important  service  which  the  study  of  thermodynamics 
renders  is  that  of  showing  what  maximum  efficiency  is  attain- 
able for  any  heat  engine  operating  under  a  given  set  of  conditions. 
It  often  happens  that  tests  indicate  an  efficiency  very  much  bet- 
ter than  is  usually  obtained  with  any  of  the  present  types  of 
engines.  In  such  cases  thermodynamic  calculations  will  show 
conclusively  whether  the  results  secured  are  possible.  The 
ability  to  interpret  correctly  the  results  of  experiments  performed 
on  all  kinds  of  heat  engines  requires  a  knowledge  of  the  basic 
principles  of  thermodynamics. 


J2  THERMQ]^Y^A^n.C  .PRINCIPLES  '  AND   DEFINITIONS 

Heat.  Heat  is  a  form  of  energy  and  not  a  material  substance. 
The  heat  of  a  body  depends  on  the  vibratory  motion  of  the  small 
particles  or  molecules  of  which  the  body  is  built  up,  the  greater 
the  velocity  and  amplitude  of  the  vibration  of  these  molecules, 
the  higher  is  the  temperature  of  the  body. 

Heat  can  be  transferred  by  conduction,  radiation  and  con- 
vection. 

The  transfer  of  heat  between  the  different  particles  of  the 
same  body  or  between  several  bodies  is  called  conduction. 

In  a  heated  body  the  particles  are  in  violent  agitation  and  as 
a  result  waves  are  formed  and  are  emitted  or  are  radiated 
through  space  to  other  bodies. 

The  transfer  of  heat  by  the  motion  of  the  heated  particles  is 
called  convection.  This  phenomenon  is  exhibited  in  liquids  and 
gases.  Thus,  when  a  hot  body  heats  the  air  in  contact  with  it, 
the  currents  of  air  which  are  produced  by  the  process  of  con- 
vection ascend  and  are  replaced  by  cooler  air. 

Temperature.  Temperature  is  the  indication  of  the  sensible 
heat  of  a  substance  and  can  be  measured  by  a  thermometer.  The 
temperature  does  not  measure  the  quantity  of  heat  energy  in  the 
substance,  but  indicates  only  the  relative  heat  intensity,  which 
can  be  revealed  by  the  senses  of  the  observer. 

Thermometers.     There  are  three  thermometric  scales: 

The  Centigrade  or  Celsius  degree  is  T^  of  the  temperature  in- 
terval between  the  melting  point  of  ice  and  the  boiling  point  of 
water  at  atmospheric  pressure,  these  two  fixed  points  being  de- 
noted o°  C.  and  100°  C.  respectively. 

The  Fahrenheit  degree  is  T|7  of  the  temperature  interval 
between  these  two  fixed  points,  the  melting  point  of  ice  being 
taken  at  32°  F.  and  the  boiling  point  of  water  at  212°  F. 

The  Reaumur  scale  has  the  melting  point  of  ice  at  o°  R.  and  the 
boiling  point  of  water  at  80°  R. 

The  thermometric  scales  generally  used  are  the  Centigrade 
and  Fahrenheit,  the  relations  between  these  scales  being: 

Degrees  C.  =  |  [degrees  F.  —  32].  (i) 

Degrees  F.  =  f  degrees  C.  -f-  32.  (2) 


j  UNITS   OF  HEAT  3 

Mercury  thermometers,  as  ordinarily  constructed,  have  the 
space  above  the  mercury  under  a  vacuum.  Such  thermometers 
cannot  be  used  for  the  measurement  of  temperatures  exceeding 
500°  F.,  as  the  vacuum  reduces  the  boiling  point  of  mercury. 
The  range  of  mercury  thermometers  can  be  increased  to  about 
900°  F.  by  filling  the  space  above  the  mercury  with  some  inert 
gas  like  nitrogen. 

For  the  measurement  of  very  high  temperatures  thermo- 
electric pyrometers  are  best  suited. 

Absolute  Zero.  In  the  graduation  of  liquid  thermometers,  the 
vaporization  of  the  liquid  at  high  temperatures  and  its  freezing 
at  low  temperatures  limits  the  thermometric  range.  The  funda- 
mental scale  of  temperature  measurement  is  based  on  Thom- 
son's absolute  thermometric  scale,  which  is  independent  of  the 
nature  of  any  thermometric  substance.  The  zero  of  the  abso- 
lute scale  or  the  absolute  zero  is  taken  as  a  point  which  marks  the 
absence  of  heat  energy  or  of  molecular  vibrations  of  a  body. 

The  absolute  zero  is  459.5°  (practically  460°)  below  the  zero 
on  the  Fahrenheit  scale  and  273.0°  below  the  zero  on  the  Centi- 
grade scale. 

Calling  the  absolute  temperature  T  and  the  temperature  as 
measured  by  a  thermometer  /, 

On  the  Fahrenheit  scale  T  =  t  +  460.  (3) 

On  the  Centigrade  scale  T  =  t  +  273.  (4) 

Units  of  Heat.  Heat  is  measured  in  heat  units.  A  heat  unit 
is  the  amount  of  heat  required  to  raise  the  temperature  of  one 
unit  weight  of  water  one  degree.  In  the  English  system  of 
measures  the  heat  unit  is  the  British  thermal  unit  (B.t.u.),  which 
is  defined  as  the  amount  of  heat  required  to  raise  the  tempera- 
ture of  one  pound  of  water  one  degree  on  the  Fahrenheit  scale. 
The  heat  unit  in  the  metric  system  is  the  calorie  *  which  is  de- 
fined as  the  heat  required  to  raise  the  temperature  of  one  kilo- 
gram of  water  one  degree  on  the  Centigrade  scale. 

*  Since  one  kilogram  =  2.204  pounds  and  one  degree  C.  =  f°F.,  i  calorie 
=  f  X  2.204  =  3-968  B.t.u. 


4  THERMODYNAMIC   PRINCIPLES  AND  DEFINITIONS 

To  correctly  define  the  British  thermal  unit  it  is  necessary 
to  state  at  what  temperature  the  rise  of  one  degree  on  the  Fahren- 
heit scale  is  to  occur,  because  the  specific  heat  of  water  is  slightly 
variable.  This  heat  unit  (B.t.u.)  is  sometimes  defined  as  the 
amount  of  heat  required  to  raise  the  temperature  of  water  one 
degree  Fahrenheit  at  the  condition  of  maximum  density  of  water, 
that  is,  between  39  and  40  degrees  Fahrenheit.  Other  definitions 
are  based  oh  the  amount  of  heat  required  to  raise  the  temperature 
of  water  one  degree  Fahrenheit  between  60  to  6 1  degrees  Fahren- 
heit. Still  another  definition,  which  is  generally  considered  to  be 
the  most  accurate,  defines  a  British  thermal  unit  as  one  one- 
hundred-and-eightieth  (y^)  of  the  amount  of  heat  required  to 
raise  the  temperature  of  water  from  32  to  212  degrees  Fahrenheit. 
In  other  words,  according  to  this  last  definition,  the  British  ther- 
mal unit  is  the  average  value  of  the  amount  of  heat  required  to 
raise  the  temperature  of  one  pound  of  water  one  Fahrenheit 
degree  between  the  conditions  of  freezing  and  boiling  at  atmos- 
pheric pressure. 

Specific  Heat.  As  the  addition  of  the  same  quantity  of  heat 
will  not,  as  a  rule,  produce  the  same  temperature  changes  in  equal 
weights  of  different  substances,  it  is  evident  that  the  amount  of 
heat  in  any  substance  will  depend  on  the  capacity  of  the  substance 
for  heat.  For  this  reason  it  is  necessary  to  allow  for  the  relative 
heat  capacity  or  the  specific  heat  (C)  of  a  substance.  Specific 
heat  can  be  defined  as  the  ratio  of  the  heat  added  to  the  tempera- 
ture change  produced  in  a  unit  weight  of  a  substance.  It  can 
also  be  defined  as  the  resistance  which  a  substance  offers  to  a 
change  in  its  temperature,  the  resistance  of  water  being  taken  as 
unity.  In  the  English  system  the  specific  heat  is  the  number  of 
British  thermal  units  (B.t.u.)  required  to  raise  the  temperature 
of  a  pound  of  the  substance  one  degree  Fahrenheit. 

Thus  if  Q  is  the  quantity  of  heat  added  to  one  pound  of  a  sub- 
stance, the  temperature  change  is, 

k.  —  h  —  7^,  (5) 

or  Q  =  C  (fe  -  /i). 


SPECIFIC  HEAT  5 

If  the  specific  heat  is  a  variable, 

:dt.  (6) 

The  following  problem  illustrates  the  application  of  equation 
(6): 
The  specific  heat  of  a  substance  is  expressed  by  the  equation, 

C  =  0.24112  +  0.000009  /. 

What  amount  of  heat  is  required  to  raise  the  temperature  of 
one  pound  of  the  substance  from  o°  to  ioo°'F.? 
Solution.    Since  the  specific  heat  is  variable, 

Q  =  Fed*. 

«4 

substituting  the  value  of  C  and  integrating, 

(0.24112  +  0.000009  /)  at 

_m 

nioo°  ,  m100' 

=  0.24112  [*L     +  0.000009 

L2Jo° 

=  0.24112  (100)  -f  0.000009  (5000) 
=  24.112  +  0.045  =  24-I57  B.t.u. 

The  specific  heat  of  gases  and  vapors  changes  considerably  in 
value  according  to  the  conditions  under  which  the  heat  is  applied. 
If  the  heat  is  applied  to  a  gas  or  a  vapor  held  in  a  closed  vessel, 
with  no  change  in  volume,  no  work  is  performed,  and,  therefore, 
all  the  heat  added  is  used  to  increase  the  temperature.  This  is 
the  condition  in  a  boiler  when  no  steam  is  being  drawn  off.  In 
this  case  the  symbol  Cv  represents  the  specific  heat  during  heat 
application  at  constant  volume.  If,  on  the  other  hand,  the  heat- 
ing is  done  while  the  pressure  is  kept  constant  and  the  volume  is 
allowed  to  change  permitting  expansion  and  the  performance  of 
work,  the  symbol  Cv  is  used  and  represents  the  specific  heat  dur- 
ing heat  application  at  constant  pressure. 


6  THERMODYNAMIC  PRINCIPLES  AND  DEFINITIONS 

When  the  problem  deals  with  w  pounds  of  a  substance  instead 
of  a  unit  weight,  equation  (6)  becomes, 


c-«jr 


(7) 


Pressure.  Force  per  unit  of  area  is  called  pressure.  Thus, 
the  pressure  exerted  by  a  gas  or  vapor  is  expressed  in  the 
English  system  in  pounds  per  square  inch,  pounds  per  square 
foot,  inches  of  mercury  or  atmospheres.  In  the  metric  system, 
pressure  is  expressed  in  kilograms  per  square  centimeter  or 
millimeters  of  mercury. 

Gages  always  read  pressures  above  atmospheric  pressure  or 
above  vacuum.  The  absolute  pressure  is  the  sum  of  the  gage 
and  atmospheric  pressures.  Thus,  if  a  gage  reads  75  pounds 
pressure  (per  square  inch),  and  the  barometer  is  29.65  inches  of 
mercury,  the  atmospheric  or  barometric  pressure  is,  in  pounds  per 
square  inch, 

29.65  X  0.491  =  14.56. 

(0.491  is  the  weight  of  a  cubic  inch  of  mercury  at  70°  F.) 
The  absolute  pressure  is: 

75  +  14.56  =  89.56  pounds  per  square  inch. 

In  thermodynamic  equations  the  unit  of  pressure  is  usually 
expressed  in  pounds  per  square  foot. 

Volume.  By  specific  volume  is  meant  the  amount  of  space 
occupied  by  a  unit  weight  of  a  substance,  expressed  in  cubic  feet 
or  in  cubic  meters.  Thus,  the  volume  of  one  pound  of  steam  at 
atmospheric  pressure  is  its  specific  volume  and  is  equal  to  26.79 
cubic  feet. 

Work.  Work  is  done  by  a  force  during  a  given  displacement 
and  is  independent  of  the  time.  The  foot-pound  is  the  unit  of 
work  in  the  English  system.  Thus,  when  a  body  weighing  one 
pound  is  raised  through  a  distance  of  one  foot,  the  resulting  work 
is  a  foot-pound.  Similarly,  the  product  of  the  pressure  in  pounds 


POWER 


per  square  foot  and  the  volume  in  cubic  feet  is  equal  to  work  in 
foot-pounds. 


0  Space  S 

FIG.  i.  —  Work  Diagram. 

Work  being  the  product  of  two  dimensions,  it  may  be  repre- 
sented graphically  by  the  area  of  a  closed  figure,  the  coordinates 
of  which  are  force  and  distance,  or  pressure  and  volume.  Thus, 
A  (Fig.  i)  represents  the  position  of  a  body  between  F  (force)  and 
S  (space)  coordinates,  0  being  the  origin  or  starting  point.  The 
distance  A  D  from  the  OS  line  represents  the  force  against  which 
it  is  acting  and  A  A',  measured  from  the  OF  line,  is  the  distance  it 
has  moved  from  the  starting  point.  If  the  body  moves  from  A 
to  B  along  the  path  AB,  the  area  A  BCD  represents  the  work 
done. 

If  the  equation  of  the  path  is  known,  the  work  is 


W 


•L 


SB 
FdS. 


(8) 


Power.  Power  is  the  rate  of  doing  work,  or  the  work  done 
divided  by  the  time  required  to  do  it.  In  the  English  system,  the 
unit  of  power  is  the  horse  power.  It  is  the  power  required  to 
raise  550  pounds  through  a  vertical  distance  of  one  foot  in  one 


8  THERMODYNAMIC  PRINCIPLES  AND   DEFINITIONS 

second,  or  33,000  pounds  one  foot  in  one  minute.  To  obtain  the 
horse  power,  the  work  in  foot-pounds  per  minute  must  be  obtained 
and  the  result  divided  by  33,000.  In  the  metric  system,  the  unit 
of  power  is  the  watt.  One  horse  power  is  equal  to  746  watts. 
The  French  horse  power  (cheval)  is  542.5  foot-pounds  per  second. 

Indicated  Horse  Power.  The  term  indicated  horse  power  is 
applied  to  the  rate  of  doing  work  by  a  gas  or  a  vapor  in  the  cylin- 
der of  an  engine.  It  is  obtained  by  means  of  an  engine  indicator. 

If  P  represents  the  mean  effective  pressure  (average  unbal- 
anced pressure)  in  pounds  per  square  inch,  as  shown  by  the 
indicator  card,  and  A  the  effective  area  of  the  piston  in  square 
inches,  the  total  pressure  exerted  on  the  piston  is  PA,  If  the 
piston  has  a  stroke  of  L  feet,  the  work  per  stroke  is  PAL  and  the 
work  per  minute  is  PALN,  where  N  represents  the  number  of 
revolutions  per  minute.  The  indicated  horse  power  is 

T  ,  Work  per  minute       PA  L  N 

I.h.p.  =  -  £-  -  =  -       —  (9) 

33>oo°  33,ooo 

Brake  Horse  Power  represents  the  actual  power  which  an 
engine  can  deliver  for  the  purposes  of  work.  The  difference 
between  indicated  and  brake  horse  power  of  an  engine  represents 
the  horse  power  lost  in  friction.  The  brake  horse  power  can  be 
measured  by  some  form  of  friction  brake,  which  absorbs  the  power 
measured,  and  is  called  an  absorption  dynamometer,  or  by  a 
transmission  dynamometer.  In  either  type  of  dynamometer, 
if  F  is  the  effective  pull  in  pounds,  L  the  lever  arm,  in  feet, 
through  which  the  weight  is  exerted,  and  N  the  number  of  revo- 
lutions of  the  shaft  per  minute,  the  brake  horse  power  is 

T,,  2TTFLN 

B.h.p.  =  --  (10) 


Mechanical  Equivalent  of  Heat.  There  is  a  definite  quantita- 
tive relation  between  work  expended  and  heat  produced.  This 
relation  between  heat  and  work  is  called  the  mechanical  equiva- 
lent of  heat  and  is  designated  by  J.  In  the  English  system, 

J  =  778  foot-pounds  - 
or  i  B.t.u.  =  778  foot-pounds. 


EFFECTS   OF  HEAT  APPLICATION  9 

In  the  metric  system, 

/  =  427  kilogrammeters 
or  i  calorie  =  427  kilogrammeters. 

The  reciprocal  of  /,  or  the  heat  equivalent  of  work,  is  designated 


by  A ,  where  A  =  — -  in  the  English  system  or  -  -  in  the  metric 

system. 

First  Law  of  Thermodynamics.  The  statement  of  the  definite 
relation  between  heat  and  mechanical  work  constitutes  what  is 
known  as  the  first  law  of  thermodynamics.  It  is  usually 
expressed : 

"Heat  and  mechanical  energy  are  mutually  convertible  and 
heat  requires  for  its  production  and  produces  by  its  disappear- 
ance mechanical  work  in  the  ratio  of  the  mechanical  equivalent 
of  heat."  In  other  words,  this  law  is  a  statement  of  the  conserva- 
tion of  energy  as  regards  the  equivalence  of  mechanical  work  and 
heat. 

Second  Law  of  Thermodynamics.  "In  order  to  transform  the 
heat  of  a  body  into  work,  heat  must  pass  from  that  body  into 
another  at  a  lower  temperature."  Thus,  there  must  be  a  dif- 
ference of  level  in  the  transformation  of  heat  energy  into  work 
and  heat  cannot  be  transformed  from  one  body  to  another  at  a 
higher  temperature,  unless  work  is  expended  in  order  to  produce 
such  a  transfer  of  heat. 

This  law  states  as  regards  heat  engines  the  limits  to  their  pos- 
sible performance,  which  would  be  otherwise  unlimited,  if  only 
the  "first  law"  of  thermodynamics  is  considered.  It  means,  also, 
that  no  heat  engine  converts  or  can  convert  into  work  all  of  the 
heat  supplied  to  it.  A  very  large  part  of  the  heat  supplied  is 
necessarily  rejected  by  the  engine  in  the  form  of  unused  heat. 

Effects  of  Heat  Application.  If  a  quantity  of  heat  Q  is  im- 
parted to  a  body,  the  following  effects  will  be  produced: 

1.  The  temperature  of  the  body  will  rise; 

2.  The  volume  of  the  body  will  increase; 

3.  The  body  will  be  capable  of  doing  external  work. 


10          THERMODYNAMIC  PRINCIPLES  AND   DEFINITIONS 

Representing  the  above  effects  by  S,  V  and  Wt 

Q=S+V  +  W,  (n) 

in  which  S  represents  that  quantity  of  heat,  termed  the  sensible 
heat,  which  was  utilized  in  raising  the  temperature  of  the  body;  V 
represents  that  quantity  of  heat  which  was  absorbed  by  the  body 
in  increasing  its  store  of  internal  energy  other  than  that  associated 
with  the  sensible  heat;  and  W  is  the  heat  which  was  absorbed  to 
perform  the  external  work  in  increasing  the  volume  of  the  body 
against  the  resistance  offered  by  external  substances.  S  +  V 
represents  "that  portion  of  the  heat  that  was  chargeable  to  the 
increase  of  internal  work.  This  may  be  termed  the  intrinsic 
energy  increase  and  designated  by  /. 
Thus,  for  the  addition  of  an  infinitesimal  quantity  of  heat, 

dQ  =  dl  +  dW 

/>/2  /»   Vt 

or  6=1   -£//+.  /      PdV.  (12) 

«//!  t/Fl 

In  equation  (12),  which  represents  the  effect  of  heat  application, 

the  intrinsic  energy  change  /    dl  depends  on  the  physical  state 

•/li 
of  the  body,  that  is,  whether  it  is  a  solid,  a  liquid,  a  vapor,  or  a 

f*  Vz 

gas,  while  the  external  work    I        P  dV  depends  on  the  char- 

J  Vi 

acter  of  the  path  in  the  work  diagram. 

The  Heat  Engine.  The  heat  engine  is  a  machine  which  con- 
verts the  heat  energy  of  solid,  liquid  or  gaseous  fuel  into  work. 
This  conversion  depends  on  the  variation  in  the  pressure,  volume 
and  temperature  of  a  gas  or  a  vapor  and  can  be  accomplished  in 
two  ways: 

First,  by  "external  combustion,"  in  which  case  the  fuel  is 
burned  outside  of  the  engine  cylinder;  the  heat  developed  by  the 
combustion  of  the  fuel  is  conducted  to  the  working  substance  or 
heat  medium  through  walls;  this  working  substance  does  work 
on  a  piston  in  the  case  of  a  reciprocating  engine  or  on  a  vane  or 


THERMAL  EFFICIENCY  OF  A  HEAT  ENGINE  II 

"  blade  "  in  a  turbine.  To  this  class  belong  steam  engines  of  the 
reciprocating,  turbine  or  rotary  types  and  also  external-combus- 
tion hot  air  engines.  Thus,  in  the  case  of  the  steam  engine,  the 
fuel,  which  may  be  coal,  wood,  petroleum  or  gas,  is  burned  out- 
side of  the  boiler  shell  and  the  resulting  heat  is  transmitted  by 
conduction  through  the  metal  of  the  shell  to  the  working  sub- 
stance, which  is  water.  When  enough  heat  has  been  added  to  the 
water  to  produce  a  change  in  its  physical  state,  water  vapor  or 
steam  is  formed  at  the  required  pressure.  This  vapor,  which 
may  be  dry,  wet  or  superheated,  if  allowed  to  act  on  the  piston 
of  the  engine,  will  do  work. 

Another  method  of  converting  heat  into  mechanical  energy  is 
by  burning  the  fuel  rapidly  or  slowly  inside  of  an  engine  cylinder 
or  in  a  communicating  vessel,  the  products  of  combustion  being 
allowed  to  act  directly  on  the  piston  of  the  engine.  To  this  class 
belong  gas,  petroleum  and  alcohol  engines  which  are  called 
"  internal  combustion  "  engines. 

Thermal  Efficiency  of  a  Heat  Engine.  By  thermal  efficiency 
(E)  is  meant  the  ratio  of  the  heat  converted  into  work  (AW)  to 
the  heat  supplied  by  the  engine  (Qi),  or 

AW 


Since  only  a  part  of  the  heat  supplied  to  an  engine  can  be  con- 
verted into  work,  the  above  ratio  is  a  fraction  always  less  than 
unity. 

PROBLEMS 

1.  Give  examples  of  the  transfer  of  heat  by  conduction,  radiation  and 
convection. 

2.  Show  that  the  kilogram  calories  per  kilogram  X  1.8  give  B.t.u.  per 
pound. 

3.  Prove  that  the  product  of  a  pressure  in  pounds  per  square  foot  and  a 
volume  in  cubic  feet  results  in  foot-pounds  of  work. 

4.  If  the  specific  heat  of  a  substance  is  0.65,  how  many  B.t.u.  are  re- 
quired to  raise  the  temperature  of  10  Ibs.  of  the  substance  through  10°  F.? 

5.  The  specific  heat  of  a  substance  is  expressed  by  the  equation, 

C  =  O.5  —  O.C2  t. 


12  THERMODYNAMIC   PRINCIPLES  AND   DEFINITIONS 

What  heat  is  required  to  raise  the  temperature  of  5  Ibs.  of  the  substance 
from  10°  to  100°  F.? 

6.  Convert  —  40°  C.  into  degrees  Fahrenheit. 

7.  Change  350°  F.,  212°  F.  and  160°  C.  to  absolute  Fahrenheit  tempera- 
tures. 

8.  Prove  that  the  weight  of  i  cu.  in.  of  mercury  is  equal  to  0.491  Ib. 

9.  If  the  barometric  reading  is  29.2  ins.,  change  140  Ibs.  per  sq.  in.  gage 
and  also  27  ins.  vacuum  into  pounds  per  square  inch  absolute  pressure. 

10.  Calculate  the  indicated  horse  power  of  a  12"  X  13"  steam  engine 
which  operates  at  a  speed  of  265  r.p.m.    The  mean  effective  pressure  of  the 
head  end  is  27.5  Ibs.  per  sq.  in.  and  of  the  crank  end  27.8  Ibs.  per  sq.  in.    The 
diameter  of  the  piston  rod  is  if  inches. 

11.  Calculate  the  brake  horse  power  developed  by  an  engine  as  measured 
by  a  Prony  brake,  the  effective  pull  being  32  Ibs.  at  250  r.p.m.  and  the 
lever  arm  32  inches  long. 

12.  One  pound  of  fuel  has  a  heating  value  of  14,500  B.t.u.    How  many 
foot-pounds  of  work  is  it  capable  of  producing,  if  all  this  heat  is  converted 
into  work? 

13.  An  engine  developed  15,560  foot-pounds  of  work.    How  much  heat 
in  B.t.u.  was  theoretically  required? 

14.  A  heat  engine  receives  100,000  B.t.u.  of  heat  in  the  form  of  fuel  and 
during  the  same  period  30,000  B.t.u.  are  converted  into  work.    What  per- 
centage (thermal  efficiency)  of  the  heat  received  by  the  engine  was  converted 
into  work? 

15.  A  gas  engine  receives  20,000  B.t.u.  of  heat  in  the  form  of  fuel  and 
during  the  same  period  3,1 12,000  foot-pounds  of  work  are  developed.    What 
is  the  thermal  efficiency  of  the  engine? 

16.  It  is  claimed  that  a  certain  motor  generates  300,000  foot-pounds  of 
work  per  hour  and  during  this  period  receives  400  B.t.u.  of  heat  in  the  form 
of  fuel.    Are  such  results  possible? 

17.  An  oil  engine  uses  0.74  Ib.  of  fuel  per  b.h.p.  per  hour.     Calculate  the 
thermal  efficiency  of  this  engine  if  the  oil  has  a  calorific  value  of  18,600  B.t.u. 
per  Ib. 

18.  An  engine  receives  200  B.t.u.  of  heat  per  minute  and  exhausts  during 
the  same  period  100  B.t.u.     If  no  losses  of  heat  occur  within  the  cylinder, 

(a)  How  many  B.t.u.  of  heat  are  being  transformed  into  work? 

(b)  What  number  of  foot-pounds  does  this  heat  produce? 

(c)  What  horse  power  is  being  developed? 

(d)  What  is  the  thermal  efficiency  of  the  engine? 

19.  In  the  manufacture  of  certain  explosives,  acids  are  mixed  with  an 
oxidizable  substance.     During  the  process  the  mixture  must  be  constantly 
agitated  by  a  stirring  mechanism  to  maintain  uniform  conditions,  and  the 


PROBLEMS  13 

temperature  of  the  mixture  must  be  kept  below  a  certain  predetermined 
value  to  prevent  explosion.  If  during  the  process  of  manufacture  50x30 
foot-pounds  of  work  are  delivered  per  minute  to  the  agitator,  and  1000  B.t.u. 
are  generated  during  the  same  period  by  the  chemical  reaction,  how  much 
heat  must  be  absorbed  per  hour  to  maintain  the  temperature  of  the  mixture 
constant? 

20.    Prove  that  one  horse  power  developed  for  one  hour  is  equivalent  to 
the  consumption  of  2545  B.t.u.  of  heat  in  the  same  period. 


CHAPTER  II 


PROPERTIES   OF  PERFECT   GASES 

In  the  study  of  thermodynamics,  the  working  substance,  or 
heat  medium,  through  which  the  heat  engine  converts  heat  into 
work,  is  in  the  condition  of  a  perfect  gas  or  vapor.  The  laws 
governing  the  action  of  the  two  classes  of  substances  differ.  For 
this  reason  the  study  of  thermodynamics  is  divided  into  the 
thermodynamics  of  gases  and  of  vapors. 

When  the  term  "  gas  "  is  used  it  refers  to  what  is  more  properly 
called  a  perfect  gas.  A  perfect  gas  may  be  denned  as  a  fluid 
which  remains  in  the  gaseous  state  when  subjected  to  changes  in 
pressure  or  in  temperature.  Gases  which  are  near  their  point 
of  condensation  are  not  perfect  gases.  Oxygen,  hydrogen,  nitro- 
gen, air  and  carbon  dioxide  are  practical  examples  of  perfect 
gases. 

Vapors  are  fluids  which  are  readily  transformed  into  liquids  by 
a  very  moderate  reduction  in  temperature  or  increase  in  pressure. 

Common   examples  of  vapors  are 
steam  and  ammonia. 

Relation  between  Pressure,  Vol- 
ume and  Temperature  of  a  Per- 
fect Gas.  In  practically  all  heat 
engines,  work  is  done  by  changes 
of  volume  of  a  fluid,  and  the 
amount  of  work  performed  depends 
only  on  the  relation  of  pressure  to 
volume  during  such  change  and 
not  at  all  on  the  form  of  the  vessel 
containing  this  fluid. 

Figure  2  shows  a  vessel  containing  a  perfect  gas  and  surrounded 
by  a  jacket  rilled  with  cracked  ice.  Its  temperature  will,  there- 

14 


Connection  to  Air  Pump 


FIG.  2.  —  Constant  Temperature 
Apparatus  for  Demonstrating 
Relation  between  Pressure  and 
Volume  of  a  Gas. 


BOYLE'S  LAW  15 

fore,  be  at  32°  F.  This  vessel  has  a  tightly  fitting  piston  P  of 
which  the  lower  flat  side  has  an  area  of  one  square  foot.  In  the 
position  shown  the  piston  is  two  feet  from  the  bottom  of  the 
vessel,  so  that  the  volume  between  the  piston  and  the  bottom  of 
the  vessel  is  two  cubic  feet.  The  pressure  on  the  gas  is  that  due 
to  the  piston  and  the  weights  shown.  Assume  this  total  weight 
is  TOO  pounds  and  that  the  air  pump  connected  to  the  top  of  the 
vessel  maintains  a  vacuum  above  the  piston.  Then  the  pressure 
on  the  gas  below  the  piston  is  100  pounds  per  square  foot.  If  now 
the  weights  are  increased  to  make  the  pressure  on  the  gas  200 
pounds  per  square  foot  the  piston  will  sink  down  until  it  is  only 
one  foot  from  the  bottom  of  the  vessel,  provided  the  ice  keeps  a 
constant  temperature.  If  the  temperature  is  not  maintained 
constant,  because  of  the  tendency  of  gases  to  expand  with  in- 
crease in  temperature,  it  will  be  necessary  to  apply  a  total  weight 
greater  than  200  pounds  to  reduce  the  volume  to  one  cubic  foot. 
Similarly,  if  the  weight  on  the  gas  were  reduced  to  50  pounds  and 
the  vessel  were  made  high  enough,  the  lower  side  of  the  piston 
would  then  be  four  feet  from  the  bottom  of  the  vessel. 

Examination  of  the  above  figures  shows  that  if  the  temperature 
is  constant  the  product  of  pressure  and  volume  is  a  constant,  and 
in  this  particular  case  it  is  always  equal  to  200  foot-pounds. 
These  facts  are  expressed  by  Boyle's  Law  which  can  be  stated  as 
follows : 

Boyle's  Law.  If  a  unit  weight  of  gas  is  compressed  or  ex- 
panded at  constant  temperature,  the  pressure  varies  inversely  as 
the  volume,  or  the  product  of  pressure  and  volume  remains  a 
constant.  Thus,  if  PI,  Vi  are  the  initial  pressure  and  volume  and 
P%,  V2  the  final  pressure  and  volume, 

PM  =  P2F2.  (14) 

The  laws  of  thermodynamics  dealing  with  volume  and  pressure 
changes  corresponding  to  temperature  variations  may  be  stated 
as  follows: 

(i)  Under  constant  pressure  the  volume  of  a  given  mass  of 
gas  varies  directly  as  the  absolute  temperature. 


16  PROPERTIES  OF  PERFECT  GASES 

(2)  Under  constant  volume  the  absolute  pressure  of  a  given 
mass  of  gas  varies  directly  as  the  absolute  temperature.  • 

These  fundamental  principles,  often  called  Gay-Lussac's  or 
Charles'  Laws,  may  also  be  stated  thus: 

With  pressure  constant,  — -  =  — -,  (15) 

Vz       J-2 
r>          y 

With  volume  constant,  — -  =  ~,  (16) 

where  FI  and  F2  are  respectively  the  initial  and  final  volumes, 
PI  and  P2  are  the  initial  and  final  absolute  pressures,  and  TI  and 
Tz  are  the  absolute  temperatures  corresponding  to  the  pressures 
and  volumes  of  the  same  subscripts. 

The  following  problem  shows  applications  of  Charles'  laws: 
A  gas  has  a  volume  of  2  cubic  feet,  a  pressure  of  14.7  pounds  per 
square  inch  absolute  and  a  temperature  of  60°  F. 

(a)  What  will  be  the  volume  of  this  -gas  if  the  temperature  is 
increased  to  120°  F.,  the  pressure  remaining  constant? 

(b)  What  will  be  the  pressure  if  the  temperature  is  increased 
as  in  (a)  but  the  volume  remains  constant? 

Solution,  (a)  Since  the  pressure  remains  constant  and  the 
substance  is  a  gas,  the  volume  varies  directly  as  the  absolute 
temperature. 

Letting  Vi  and  Ti  be  the  initial  conditions  and  F2  and  Tz  be 
the  final  conditions,  then 

Fi  _  Ti  _2_       60  +  460 

"_ '~    ••"•     __~        OJL         '_ _.~    t~~ 


T2  F2       120  +  460' 

F2  =  2.23  cubic  feet. 

(b)  Since  the  volume  remains  constant, 

Pi  =  Ti  14-7        60  +  460 

p2  "  r2         P2      120 


P2  =  16.39  pounds  per  square  inch  absolute. 

Combination  of  Boyle's  and  Charles'  Laws.     Equations  (14), 
(15),  and  (16)  cannot  often  be  used  as  they  stand,  because  it 


COMBINATION  OF  BOYLE'S  AND   CHARLES'  LAWS  17 

does  not  often  happen  that  any  one  of  the  three  variables  (P,  V 
and  T)  remains  constant.  A  more  general  law  must  be  developed, 
therefore,  allowing  for  variations  in  all  of  the  terms  P,  V  and  T. 
This  is  accomplished  by  combining  the  above  equations. 

Assume  a  pound  of  gas  of  which  the  initial  conditions  of  pres- 
sure, volume  and  temperature  are  represented  by  P1}  Vi  and  TI, 
while  the  corresponding  final  conditions  are  given  by  P2,  F2  and 
TV  The  first  step  is  in  changing  the  volume  from  V\  to  F2  and 
the  pressure  from  PI  to  some  intervening  pressure  P2'  while  the 
temperature  TI  remains  constant.  This  change  can  be  expressed 
by  Boyle's  law  (equation  14). 

With  constant  temperature  (7\), 


from  which,  by  solving, 


where  P2'  is  the  resulting  pressure  of  the  gas  when  its  volume  is 
changed  from  Fi  to  F2,  with  the  temperature  remaining  constant 
at  TI. 

The  second  step  is  in  the  change  in  pressure  from  P2'  to  P2  and 
in  temperature  from  TI  to  T2,  while  the*  volume  remains  con- 
stant at  F2.  This  step  is  expressed  as  follows: 

With  constant  volume  (F2), 

P/       Ti 


which  may  be  written 

A  =  F~  (*>) 

JL  1 

Substituting  now  the  value  of  P2'  from  (18)  in  (20),  we  have 


18  PROPERTIES  OF  PERFECT  GASES 

which  may  be  arranged  to  read, 


rrr\  \          / 

1  -t  2 

The  following  problem  shows  the  application  of  equation  (22): 
A  quantity  of  air  at  atmospheric  pressure  has  a  volume  of 
2000  cubic  feet  when  the  barometer  reads  28.80  inches  of  mercury 
and  the  temperature  is  40°  C.  What  will  be  the  volume  of  this 
air  at  a  temperature  of  o°  C.  when  the  barometer  reads  29.96 
inches  of  mercury? 

Solution.    Volume,  pressure  and  temperature  vary  in  this  case 
as  in  the  following  equation, 


Letting  PI,  Fi,  TI  .=  initial  conditions, 

Pz,  F2,  T2  =  final  conditions, 

28.80  X  2000       29.96  X  F2 
then  -  =  -  —  > 

40  +  273  o  +  273 

F2  =  1676  cubic  feet. 

Now,  since  P2,  F2  and  T2  in  equation  (22)  represent  any 
simultaneous  condition  of  the  gas,  we  may  also  write  the  follow- 
ing more  general  relations : 

^^M^  a  constant,.    '  (.3) 

J-1  J-2  Jf  3 

and,  therefore, 

PV  =  RT,  (24) 

PV 

where  R  is  the  "  gas  constant,"  and  is  equal  to  -—-• 

The  Law  of  Perfect  Gases.  Equation  (24)  expresses  the  law 
connecting  the  relation  between  pressure,  volume  and  tempera- 
ture of  a  perfect  gas.  In  this  equation  F  is  the  specific  volume, 
or  the  volume  occupied  by  a  unit  weight  of  a  gas  at  the  absolute 
pressure  P  and  absolute  temperature  T;  R  is  the  gas  constant 


THE  LAW   OF   PERFECT   GASES  ig 

in  foot-pounds  and  depends  on  the  density  of  the  gas  and  on  the 
units  of  measurement  adopted. 

By  means  of  equation  (24)  if  the  pressure,  volume  and  tem- 
perature of  a  gas  for  one  given  condition  are  known,  the  value  of 
R  can  be  determined. 

Example  i.  If  the  volume  of  air  at  freezing  point  and  atmos- 
pheric pressure  is  12.39  cubic  feet  per  pound,  calculate  the  value 
of  R  in  the  English  units. 

Solution.    R  =  *f*  =  I4'7  X  '"  *  I2'39  =  53-3-. 

TQ  32  +  460 

Example  2.     If  one  pound  of  air  occupies  5  cubic  feet  at  a 
temperature  of  200°  F.,  find  the  corresponding  pressure. 
Solution.    Using  the  value  of  R  for  air  as  calculated  above 


n  \          53.3  (460  +  20O) 

Pi  =  -—  -  =  ^  -=  7040  pounds 

per  square  foot  absolute,  which  is  the  same  as  34.2  pounds  per 
square  inch  gage  pressure. 

In  calculating  R  by  equation  (24)  care  must  be  taken  not  to 
confuse  the  units  of  measurement.  It  must  also  be  remembered 
that  the  method,  as  illustrated  in  the  above  examples,  gives  the 
value  of  R  for  one  pound  of  air,  or  for  one  unit  weight  of  the  gas 
in  question;  for  w  pounds,  the  value  of  the  constant  is  wR,  or: 

PV  =  wRT,  (25) 

where  P  =  absolute  pressure  in  pounds  per  square  foot, 
V  =  volume  in  cubic  feet, 
w  =  weight  of  gas  in  pounds, 
R  =  the  "gas  constant"  for  one  pound  of  gas  in  foot- 

pound units, 
T  =  the  absolute  temperature  in  Fahrenheit  degrees. 

This  equation  is  applicable  to  any  perfect  gas  within  the  limits 
of  pressure  and  temperature  employed  in  common  engineering 
practice.  The  "  thermodynamic  "  state  of  a  gas  is  known  when 
its  pressure,  volume,  temperature,  weight  and  composition  are 
known;  when  any  four  of  these  quantities  are  known  the  fifth  can 
be  found  by  equation  (25). 


20  PROPERTIES   OF  PERFECT  GASES 

Heat  and  Its  Effect  Upon  a  Gas.  In  equation  (12)  it  was 
shown  that  in  general  the  effect  of  adding  heat  upon  a  substance 
was  to  increase  the  intrinsic  energy  and  to  overcome  the  external 
resistances  producing  work.  This  law  can  be  stated  as  follows: 
Heat  supplied  =  increase  in  intrinsic  energy  +  the  external 
work  done. 

The  converse  of  this  statement  is  equally  true.  The  heat 
abstracted  from  a  gas  equals  the  decrease  in  intrinsic  energy  plus 
the  negative  work  done. 

External  Work.  The  external  work  or  the  work  done  by  a 
gas  in  its  expansion  is  represented  graphically  by  Fig.  3.  The 

area  under  the  expansion  line  BC 
is  proportional  to  the  work  done 
~=2'  in  the  expansion.  If  the  initial 
condition  of  the  gas  at  B  as  re- 
gards pressure  and  volume  is  rep- 
resented by  PI  and  FI  and  the 
final  condition  at  C  by  PI  and  F2 
(expansion  being  at  constant  pres- 
sure), the  force  P  moved  through 

a  distance  of  abscissas  (F2  —  FI) 
FIG.  3.  —  External  Work  of  Expansion.  • 


Obviously  the  area  under  the  line  BC  divided  by  the  horizontal 
length  (F2  —  FI)  is  the  average  value  of  the  force  P.  If,  further, 
and  in  general,  we  represent  the  area  under  BC  by  the  symbol  A , 

then  we  can  write, 

^4 

_-  '          =  average  value  of  P, 
1/2  —  Ki 

whether  or  not  P  is  constant.    And  also, 

Work  done  =  T7  A  „   X  (F2  -  Fi)  =  A. 
Vz  —  KI 

The  same  principle  applies  whether  the  line  BC  is  a  straight  line, 
as  shown  in  Fig.  3,  or  a  very  irregular  curve,  as  will  be  shown 
later. 


INTERNAL   ENERGY  21 

Internal  Energy.  The  heat  energy  possessed  by  a  gas  or 
vapor,  or  the  heat  energy  which  is  contained  in  a  gas  or  vapor 
in  a  form  similar  to  "  potential "  energy,  is  called  its  internal 
energy.  It  is  also  called  intrinsic  energy,  since  it  may  be  said  to 
"reside"  within  the  substance  and  has  not  been  transferred 
to  any  other  substance.  Thus,  an  amount  of  heat  added  to  a 
substance  when  no  work  is  performed  is  all  added  to  the  internal 
energy  of  that  substance.  On  the  other  hand,  when  heat  is  added 
while  work  is  being  performed,  the  internal  energy  is  increased 
only  by  the  difference  between  the  heat  added  and  the  work 
done. 

Internal  energy  may  also  be  denned  as  the  energy  which  a  gas 
or  vapor  possesses  by  virtue  of  its  temperature,  and  for  one 
pound  of  gas  may  be  expressed  as  follows: 

Internal  energy  =  Cv  dT  (in  B.t.u.), 

where  T  is  the  absolute  temperature  and  CP  the  specific  heat  at 
constant  volume.  From  the  paragraph  on  specific  heat  it  will  be 
remembered  that  Cv  takes  into  account  only  that  heat  required 
to  raise  the  temperature,  since  under  constant  volume  conditions 
no  external  work  is  done;  and,  therefore,  in  dealing  with  internal 
energy,  C,  is  always  used. 

Increase  in  internal  energy  in  B.t.u.  (for  one  pound  of  a  gas) 

=  C,  (T2  -  2\).  (26) 

Joule's  Law.  In  the  case  of  ideally  perfect  gases,  such  as 
thermodynamic  equations  must  deal  with,  it  is  assumed,  when 
a  gas  expands  without  doing  external  work  and  without  taking 
in  or  giving  out  heat  (and,  therefore,  without  changing  its  stock 
of  internal  energy) ,  that  its  temperature  does  not  change.  It  was 
for  a  long  time  supposed  that  when  a  gas  expanded  without  doing 
work,  and  without  taking  in  or  giving  out  heat,  that  its  tempera- 
ture did  not  change.  This  fact  was  based  on  the  famous  experi- 
ments of  Joule.  Later  investigations  by  Lord  Kelvin  and  Linde 
have  shown  that  this  statement  is  not  exactly  correct  as  all  known 
gases  show  a  change  in  temperature  under  these  conditions.  This 
change  in  temperature  is  known  as  the  "  Joule-Thomson"  effect. 


22  PROPERTIES   OF  PERFECT   GASES 

Relation  of  Specific  Heats  and  the  Gas  Constant.     If  heat  is 
added  at  constant  pressure,  then 

Q  =  wC9(Ti-Tj.  (27) 

Also,  by  equation  (26),  the  increase  in  internal  energy  when 
heat  is  added 

=  we.  (r,  -  J\).  (28) 

External  work  =  P  (V*  —  Fi)  foot-pounds 


Since  the  heat  added  =  increase  in  internal  energy  +  external 
work, 

wcp  (Tt  -  ro  =  we,  (Tt  -  ro  +  F  (F*~  Fl)-      (3°) 

By  equation  (25), 

P2F2  =  ivRTs  and  P1V1  = 
Substituting  these  values  in  (30), 

we,  (Tt  -  ro  =  we.  (r,  -  ro  +  w  R^~ 

K  > 

Simplifying,  C,  =  C.  +  —  •  (32) 

(33) 

Equation  (33)  shows  that  the  difference  between  the  two 
specific  heats  is  equal  to  the  gas  constant  R,  which  when  measured 
irr  foot-pounds  represents  the  external  work  done  by  one  pound 
of  a  gas  when  its  temperature  is  increased  by  one  degree  under 
constant  pressure. 

Example.  The  specific  heat  of  air  at  constant  pressure  (Cp) 
is  0.2375  B.t.u.  and  R  =  53.3  foot-pounds.  Calculate  the  spe- 
cific heat  at  constant  volume  (C). 

Solution. 

0.2375  -^~  =  0.1690  B.t.u. 

778 


RATIO  OF  THE  TWO  SPECIFIC  HEATS  23 

Ratio  of  the  Two  Specific  Heats  ( — -V  The  constant  repre- 
senting the  ratio  of  the  two  specific  heats  of  a  perfect  gas  is 
represented  by  7  where 

=  Cp  =         CP  i 

7  ~  CD'~        _AR~        _  AR 

\*S  ft  s~*  I  x-* 


Example.  Calculate  the  value  of  7  for  air.  Cp  =  0.2375  B.t.u. 
R  =  53.3  foot-pounds. 

Solution.     7  = —  =  =  1.405. 

AR  m.i, 

i i 

Cp  778  X  0.2375 

The  constant  designating  the  relation  between  the  specific 
heats  of  air  is  often  designated  by  k  instead  of  7. 

Values  of  the  Specific  Heats.  Regnault  after  carrying  on 
experiments  on  the  specific  heat  of  hydrogen,  oxygen,  air  and 
carbon  dioxide  concluded  that  all  gases  have  a  constant  specific 
heat  under  varying  pressures  and  temperatures.  Recent  experi- 
ments tend  to  show  that  the  specific  heats  of  substances  vary 
with  the  pressure  and  temperature.  The  variability  in  the  values 
of  the  specific  heats  does  not  influence  to  any  very  great  extent 
most  thermodynamic  computations. 

In  the  application  of  thermodynamics  to  internal  combustion 
engines  the  exact  values  of  the  specific  heats  are  of  considerable 
importance. 

Tables  i  and  2  in  the  appendix  give  constants  for  various  gases. 


PROBLEMS 

1.  Air  at  constant  pressure  with  an  initial  volume  of  2  cu.  ft.  and  temper- 
ature of  60°  F.  is  heated  until  the  volume  is  doubled.    What  is  the  resulting 
temperature  in  degrees  Fahrenheit? 

2.  Air  is  cooled  at  constant  volume.    The  initial  pressure  is  30  Ibs.  per 


24  PROPERTIES  OF  PERFECT  GASES 

sq.  in.  absolute  and  the  initial  temperature  is  101°  F.    The  final  condition 
has  a  temperature  of  50°  F.    What  is  the  final  pressure? 

3.  One  pound  of  hydrogen  is  cooled  at  constant  pressure  from  a  volume 
of  i  cu.  ft.  and  temperature  of  300°  F.  to  a  temperature  of  60°  F.    What  is 
the  resulting  volume? 

4.  A  tank  whose  volume  is  50  cu.  ft.  contains  air  at  105  Ibs.  per  sq.  in. 
absolute  pressure  and  temperature  of  80°  F.    How  many  pounds  of  air  does 
the  tank  contain? 

5.  An  automobile  tire  has  a  mean  diameter  of  34  inches  and  a  width  of 
4  inches.    It  is  pumped  to  80  Ibs.  per  sq.  in.  gage  pressure  at  a  temperature 
of  60°  F.;  atmospheric  pressure  14.6  Ibs.  per  sq.  in.  absolute. 

(a)  How  many  pounds  of  air  does  the  tire  contain? 

(b)  Assuming  no  change  of  volume,  what  would  be  the  gage  pressure 
of  the  tire  if  placed  in  the  sun  at  100°  F.? 

6.  A  gas  tank  is  to  be  made  to  hold  0.25  Ib.  of  acetylene  when  the  pres- 
sure is  250  Ibs.  per  sq.  in.  gage,  atmospheric  pressure  14.4  Ibs.  per  sq.  in. 
absolute,  and  the  temperature  of  the  gas  70°  F.    What  will  be  its  volume  in 
cubic  feet? 

7.  A  quantity  of  air  at  a  temperature  of  70°  F.  and  a  pressure  of  15  Ibs. 
per  sq.  in.  absolute  has  a  volume  of  5  cu.  ft.     What  is  the  volume  of  the 
same  air  when  the  pressure  is  changed  at  constant  temperature  to  60  Ibs. 
per  sq.  in.  absolute? 

8.  How  many  pounds  of  air  are  required  for  the  conditions  in  problem  7? 

9.  The  volume  of  a  quantity  of  air  is  10  cu.  ft.  at  a  temperature  of  60°  F. 
when  the  pressure  is  15  Ibs.  per  sq.  in.  absolute.     What  is  the  pressure  of 
this  air  when  the  volume  becomes  60  cu.  ft.  and  the  temperature  60°  F.? 

10.  How  many  pounds  of  air  are  required  for  the  conditions  in  problem  9? 

11.  A  tank  contains  200  cu.  ft.  of  air  at  a  temperature  of  60°  F.  and  under 
a  pressure  of  200  Ibs.  per  sq.  in.  absolute. 

(a)  What  weight  of  air  is  present? 

(b)  How  many  cubic  feet  will  this  air  occupy  at  14.7  Ibs.  per  sq.  in. 
absolute  and  at  a  temperature  of  100°  F.? 

12.  The  volume  of  a  quantity  of  air  at  70°  F.  and  at  a  pressure  of  14.2  Ibs. 
per  sq.  in.  absolute  is  20  cu.  ft.     What  is  the  temperature  of  this  air  when 
the  volume  becomes  5  cu.  ft.  and  the  pressure  80  Ibs.  per  sq.  in.  absolute? 

13.  If  the  specific  heat  of  carbon  dioxide  under  constant  pressure  Cv  is 
0.2012  and  the  value  of  R  is  35.10,  find  the  value  of  the  specific  heat  under 
constant  volume  CP. 

14.  How  many  B.t.u.  are  required  to  double  the  volume  of  one  pound  of 
air  at  constant  pressure  from  50°  F. 

15.  A  tank  filled  with  200  cu.  ft.  of  air  at  15  Ibs.  per  sq.  in.  absolute  and 
60°  F.  is  heated  to  150°  F. 


PROBLEMS  25 

(a)  What  will  be  the  resulting  air  pressure  in  the  tank? 

(b)  How  many  B.t.u.  will  be  required  to  heat  the  air? 

1 6.  A  tank  contains  200  cu.  ft.  of  air  at  60°  F.  and  40  Ibs.  per  sq.  in.  abso- 
lute. If  $00  B.t.u.  of  heat  are  added  to  it,  what  will  be  the  resulting  pres- 
sure and  temperature? 


CHAPTER  III 
EXPANSION  AND   COMPRESSION   OF  GASES 

The  equation  of  the  perfect  gas  in  the  form  PV  =  wRT  for  the 
expansion  or  compression  of  gases  has  three  related  variables, 
(i)  pressure,  (2)  volume  and  (3)  temperature.  For  a  given 
weight  of  gas  with  any  two  of  these  variables  known  the  third  is 
fixed.  As  regards  the  analysis  of  the  action  of  heat  engines,  the 
pressure  and  volume  relations  are  most  important,  and  graphical 
diagrams,  called  pressure- volume  or  P-V  diagrams,  are  frequently 
needed  to  assist  in  the  analysis.  The  indicator  diagram  is  a  pres- 
sure-volume diagram  drawn  autographically  by  the  mechanism 
of  an  engine  indicator.  A  pressure- volume  diagram  is  shown 
in  Fig.  4,  in  which  the  vertical  scale  of  coordinates  represents 

pressures    and    the    horizontal, 


100 


volumes.  Assume  that  the  pres- 
sure  and  volume  of  a  pound  of  a 
gas  are  given  by  the  coordinates 
P  and  FI,  which  are  plotted  in 


i  the  middle  of  the  diagram.     It 


**  2       4       e       s      ic         will  be    assumed    further    that 

Volume,  eu.  ft. 

the  pressure  remains  constant  in 

FIG.  4.  —  Diagram  of  Expansion  and      ,11  i         •    j  •  i 

Compression  at  Constant  Pressure.        &*    changes     to    be     "^ated. 

Now  if  the  gas  is  expanded  until 

its  volume  becomes  V^  then  its  condition  as  regards  pressure 
and  volume  would  be  represented  by  P  F2.  If,  on  the  other  hand, 
the  gas  had  been  compressed  while  a  constant  pressure  was 
maintained,  its  final  condition  would  be  represented  by  the  point 
PVz  to  the  left  of  PV\.  Similarly,  any  line  whether  straight 
or  curved  extending  from  the  initial  condition  of  the  gas  at  PVi 
will  represent  an  expansion  when  drawn  in  the  direction  away 

26 


EXPANSION  AT  CONSTANT  PRESSURE  27 

from  the  zero  of  volumes  and  will  represent  a  compression  when 
tending  toward  the  same  zero. 

It  has  been  shown  (page  20)  that  areas  on  such  diagrams  repre- 
sent the  product  of  pressure  and  volume,  and,  therefore,  work  or 
energy.  Thus  in  Fig.  4  the  area  under  the  curve  PV3  to  PV2 
represents  on  the  scales  given  100  (pounds  per  square  foot)  X 
(9  —  1)  cubic  feet  or  800  foot-pounds  irrespective  of  whether  it  is 
an  expansion  or  a  compression  from  the  initial  condition. 

Most  of  the  lines  to  be  studied  in  heat  engine  diagrams  are 
either  straight  or  else  they  can  be  exactly  or  approximately 
represented  by  an  equation  in  the  form 

PVn  =  a  constant,  (35) 

where  the  index  n,  as  experimentally  determined,  has  varying 
numerical  values,  but  is  constant  for  any  one  curve.  When 
the  lines  of  the  diagram  are  straight  the  areas  of  simple  rectangles 
and  triangles  need  only  be  calculated  to  find  the  work  done. 
The  two  most  common  forms  of  curves  to  be  dealt  with  in  ex- 
pansions are  (i)  when  there  is  expansion  with  addition  of  heat 
at  such  a  rate  as  to  maintain  the  temperature  of  the  gas  con- 
stant throughout  the  expansion.  Such  an  expansion  is  called 
isothermal.  The  other  important  kind  of  expansion  (2)  occurs 
when  work  is  done  by  the  gas  without  the  addition  or  abstraction 
of  heat.  To  do  this  work  some  of  the  internal  heat  energy 
contained  in  the  gas  must  be  transformed  in  proportion  to  the 
amount  of  work  done.  Such  an  expansion  is  called  adiabatic. 

The  following  problems  show  the  application  of  the  foregoing 
principles  to  various  types  of  expansions: 

i.  Expansion  at  Constant  Pressure.  One  pound  of  air  having 
an  initial  temperature  of  60°  F.  is  expanded  to  100°  F.  under 
constant  pressure.  Find 

(a)  External  work  during  expansion; 

(b)  Heat  required  to  produce  the  expansion. 

Solution.  The  heat  added  equals  the  increase  in  internal 
energy  plus  the  external  work  done.  In  solving  for  the  heat 
added  or  required  during  any  expansion  it  is  only  necessary  to 


28  EXPANSION  AND   COMPRESSION  OF   GASES 

find  the  external  work  (which  is  equal  to  the  area  under  the  ex- 
pansion curve)  and  add  to  it  the  heat  needed  to  increase  the 
internal  energy. 

The  external  work  =  W  =  PI  (7,  -  Fi).  (36) 

Its  equivalent  is:  wR,(Tz  -  TI}.  (37) 

W  =  i  X  53.3  [(100  +  460)  -  (60  +  460)]  =  2i32ft.-lbs. 

The  increase  in  internal  energy: 

/  =  wcv  (TI  -  ro  (38) 

/  =  i  X  0.169  [(100  +  460)  —  (60  +  460)] 
=  6.76  B.t.u. 


Heat  required  (Q)  =  6.75  +         -  =  9.50  B.t.u. 

778 

Another  method  of  computing  the  heat  required  to  produce 
the  expansion  is: 

Q  =  wCP(T,-Tl).  (39) 

Q  =  i  x  0.237  [(100  +  460)  -  (60  +  460)] 
=  9.50  B.t.u.  approximately. 

2.  Expansion  at  Constant  Volume.     One  pound  of  air  having 
an  initial  temperature  of  60°  F.  is  heated  at  constant  volume  until 
the  final  temperature  is  120°  F.     Find: 

(a)  External  work; 

(b)  Heat  required. 

Solution. 

Heat  added  (Q)  =  increase  in  internal  energy  +  external  work. 

External  work    =  o. 

Then 

Heat  added  =  increase  in  internal  energy  +  o 
=  wCv  (r2  -  Ti)  +  o 

=  i  X  0.169  [(100  +  460)  —  (60  +  460)]  +  o 
=  6.76  B.t.u. 

3.  Expansion   and   Compression   at   Constant   Temperature 
(Isothermal).    In  an  isothermal  expansion  or  compression  the 


ISOTHERMAL  EXPANSION  AND   COMPRESSION 


29 


temperature  of  the  working  substance  is  kept  constant  through- 
out the  process.  The  form  of  the  isothermal  curve  on  pressure- 
volume  coordinates  depends  upon  the  substance.  In  the  case 
of  perfect  gases  Boyle's  Law  (equation  14)  applies  and  we  have: 


PV  =  C  =  a  constant. 


(40) 


Equation  (40)  is  that  of  a  rectangular  hyperbola.  It  is  the 
special  case  of  the  general  equation  PVn  =  constant  (35),  in 
which  the  index  n  =  i  and  is  represented  by  Fig.  5. 


P,V 


Volume  dv 

FIG.  5.  —  Work  done  during  Isothermal  Expansion  and  Compression. 

The  external  work  performed  is  shown  graphically  by  the 
shaded  area  under  the  curve  between  A  and  B  (Fig.  5).  The 
two  vertical  lines  close  together  in  the  figure  are  the  limits  of  a 
narrow  closely  shaded  area  and  indicate  an  infinitesimal  volume 
change  dV.  Work  done  during  this  small  change  of  volume  is, 
the 

dW  =  PdV, 

and  for  a  finite  change  of  volume  of  any  size  as  from  V\  to  F2 
the  work  done,  W  (foot-pounds),  is: 


W 


=r 

Jv\ 


PdV. 


(41) 


30  EXPANSION  AND   COMPRESSION  OF   GASES 

For  the  integration  of  this  form  it  is  necessary  to  substitute  P 
in  terms  of  V.  Assume  that  P  and  V  are  values  of  pressure 
and  volume  for  any  point  on  the  curve  of  expansion  of  a  gas  of 
which  the  equation  is 

PV  =  C. 

Then  P  =  ^  • 

Substituting  this  value  of  P  in  equation  (41), 

f*V2  r* 
W  =          ^dV  = 

TF  =  C(logeF2  -logeFO.  (42) 

Since  the  initial  conditions  of  the  gas  are  PI  and  FI,  we  have 

and  substituting  this  value  of  C  in  equation  (42),  we  obtain 
W  =  PiFx  (log,  F2  -  loge  70 

or  W  =  PiFi  loge  -^  (in  foot-pounds).  (43) 

Units  of  weight  do  not  enter  in  equations  (42)  and  (43).  For 
a  certain  weight  of  gas  under  the  same  conditions,  since 

Fo      P 
PiFi  =  wRT  (in  foot-pounds)  and  -=^  =  —  > 

then  the  work  for  w  pounds  is: 

F  P 

W  =  wRT  loge  ~  =  wRT  loge  •==  (in  foot-pounds).       (44) 

Often  the  ratio  •=?  is  called  the  ratio  of  expansion  and  is  rep- 
resented by  r.  Making  this  substitution  we  have, 

W  =  wRT  loge  r.  (45) 

Equations  (42),  (43),  (44)  and  (45)  refer  to  an  expansion  from 
PiFi  to  PzV-i.  If,  on  the  other  hand,  the  work  done  is  the  result 
of  a  compression  from  PiFi  to  P3F3  the  curve  of  compression 


ISOTHERMAL  EXPANSION  AND  COMPRESSION  31 

would  be  from  A  to  C  and  the  area  under  it  would  be  its  graphical 
representation.  Equations  (43),  (44)  and  (45)  would  represent 
the  work  done  for  compression  the  same  as  for  expansion,  except 
that  the  expression  would  have  a  negative  value;  that  is,  work  is 
to  be  done  upon  the  gas  to  decrease  its  volume. 

The  isothermal  expansion  or  compression  of  a  perfect  gas 
causes  no  change  in  its  stock  of  internal  energy  since  the  temper- 
ature T  is  constant.  During  such  an  expansion  the  gas  must 
take  in  an  amount  of  heat  just  equal  to  the  work  it  does,  and 
conversely  during  an  isothermal  compression  it  must  reject  an 
amount  of  heat  just  equal  to  the  work  spent  upon  it.  This 
quantity  of  heatQ  (in  B.t.u.)  is,  from  equation  (45), 

wRT  .       V2 

(46) 


The  following  problem  shows  the  application  of  the  foregoing 
formulas  to  isothermal  expansions: 

Air  having  a  pressure  of  100  pounds  per  square  inch  absolute 
and  a  volume  of  i  cubic  foot  expands  isothermally  to  a  volume 
of  4  cubic  feet.  Find 

(a)  External  work  of  the  expansion; 

(b)  Heat  required  to  produce  the  expansion; 

(c)  Pressure  at  end  of  expansion. 

Solution,     (a)  Since  the  expansion  is  isothermal, 

External  work,  W  =  Px7i  loge*  ^ 

=  100  x  144  x  i  x  1.3848 
=  19,940  foot-pounds. 

(b)  Since  the  heat  added  equals  increase  in  internal  energy 
plus  external  work,  and  since  the  temperature  remains  constant 
(requiring  therefore  no  heat  to  increase  the  internal  energy),  the 
internal  energy  equals  zero  and  the  heat  added  equals  the  work 
done. 

*  2.3  X  log  base  10  =  log  base  e.  Tables  of  natural  logarithms  are  given  in  the 
appendix. 


EXPANSION  AND   COMPRESSION  OF   GASES 


Then:      Heat  added  =  external  work  = 


778 


=  25.6  B.t.u. 


(c)  Since  PlVi  =  P2F2, 

then  100  X  i  =  PZ  X  4, 

P2  =  25  pounds  per  square  inch  absolute. 

If  a  gas  expands  and  does  external  work  without  receiving  a 
supply  of  heat  from  an  external  source,  it  must  derive  the  amount 
of  heat  needed  to  do  the  work  from  its  own  stock  of  internal 
energy.  This  process  is  then  necessarily  accompanied  by  a  low- 
ering of  temperature  and  the  expansion  obviously  is  not  iso- 
thermal. 

4.  Adiabatic  Expansion  and  Compression.  In  the  adiabatic 
mode  of  expansion  or  compression  the  working  substance  neither 
receives  nor  rejects  heat  as  it  expands  or  is  compressed.  A 
curve  which  shows  the  relation  of  pressures  to  volumes  in  such 
a  process  is  called  an  adiabatic  line  (see  Fig.  6).  In  any  adiabatic 


Volume 
FIG.  6. —  Isothermal  and  Adiabatic  Expansion  Lines. 

process  the  substance  is  neither  gaining  nor  losing  heat  by 
conduction,  radiation  or  internal  chemical  action.  Hence  the 
work  which  a  gas  does  in  such  an  expansion  is  all  done  at  the 
expense  of  its  stock  of  internal  energy,  and  the  work  which  is 
done  upon  a  gas  in  such  a  compression  all  goes  to  increase  its 
internal  energy.  Ideally  adiabatic  action  could  be  secured  by 
a  gas  expanding,  or  being  compressed,  in  a  cylinder  which  in  all 
parts  was  a  perfect  non-conductor  of  heat.  The  compression  of 
a  gas  in  a  cylinder  is  approximately  adiabatic  when  the  process  is 
very  rapidly  performed,  but  when  done  so  slowly  that  the  heat 


ADIABATIC   EXPANSION  AND  COMPRESSION  33 

has  time  to  be  dissipated  by  conduction  the  compression  is  more 
nearly  isothermal.  Fig.  6  shows  on  a  pressure- volume  diagram 
the  relation  between  an  isothermal  and  an  adiabatic  form  of  ex- 
pansion or  compression  from  an  initial  condition  P\V\  at  A  to 
final  conditions  at  B  and  C  for  expansions,  and  at  D  and  E  for 
compressions. 

In  order  to  derive  the  pressure-volume  relation  for  a  gas 
expanding  adiabatically,  consider  the  fundamental  equation 
(page  20): 

Heat  added  =  increase  in  internal  energy  +  external  work,  or : 

Q  =  ivKv  (T2  -  Ti)  +  P  dV  (foot-pounds),  (47) 

where  Kv  is  the  specific  heat  in  foot-pound  units,  i.e.,  778  Ct. 
In  the  adiabatic  expansion  no  heat  is  added  to  or  taken  away 
from  the  gas  by  conduction  or  radiation,  and,  therefore,  the  left 
hand  member  of  the  above  equation  becomes  zero.  Furthermore, 
since  equation  (25)  can  always  be  applied  to  perfect  gases,  the 
following  simultaneous  equations  may  be  written: 


o  =  wKv  dT  +  P  dV,  (48) 

PV  =  wRT.  (49) 

When  P,  V  and  T  vary,  as  they  do  in  adiabatic  expansion, 
equation  (49)  may  be  written  as  follows: 

PdV+  VdP  =  wRdT,  (50) 

and 

dT  -PdV  +  VdP 
wR 

Substituting  the  value  of  dT  in  (48), 

PdV  +  VdP 

-  =  o.  (51) 


RP  dV  +  K,P  dV  +  KEV  dP  =  o. 


34  EXPANSION  AND   COMPRESSION  OF   GASES 

To  separate  the  variables  divide  by  PV: 

R^Y.jLK—  +  K  —  =  o 
Collecting  terms, 


Integrating, 

(R  +  Kv]  loge  V  +  Kv  \oge  P  =  a  constant  =  c.        (53) 


R  +  Kv 

K°       =C,  (S4) 

since  from  equation  (33), 

R  +  Kv  =  KPJ 

where  Kp  and  Kv  are  respectively  the  specific  heats  at  constant 
pressure  and  at  constant  volume  in  foot-pound  units,  then  equa- 
tion (54)  becomes: 


From  equation  (34), 


\QgeP  VKV  =  c. 


CD  Kp 

gr  =  7,     or     —  =  T, 

therefore : 

PVy  =  a  constant.  (55) 

Following  the  method  used  for  obtaining  an  expression  for 
the  work  done  in  isothermal  expansion  (equation  43),  the  work 
done,  W  (in  foot-pounds),  for  a  change  of  volume  from  Vi  to  F2, 

W  =  f^P  dV.  (56) 

For  purposes  of  integration,  P  can  be  substituted  in  terms  of 
V  as  outlined  below.  In  the  general  expression  PVn  =  c,a,  con- 
stant (see  equation  35),  where  P  and  V  are  values  of  pressure  and 
volume  for  any  point  on  the  curve  of  expansion  of  a  gas  of  which 


ADIABATIC  EXPANSION  AND  COMPRESSION  35 

the  initial  condition  is  given  by  the  symbols  PI  and  FI,  we  can 
then  write, 

P  =  f»'  (57) 

And  substituting  (57)  in  (56), 

f*v*  r  f*v*  flV 

w  -£$*-<£%•        (s8) 

-n+l     -F2 


nv-»  -  F!1-"!  ,  -. 

=  CL     x-»    J"  (59) 

Since  PFn  =  c  =  PiFin  =  P2F2n,  we  can  substitute  for  c  in 
(59)  the  values  corresponding  to  the  subscripts  of  F  as  follows: 


w 


P2F2nF2J-n  -  PiFiTi1 


i  —  n 

P.T7-  ._    I 

w  = 


P2F2- 


i  —  n 


or  W  =  — !— —  (foot-pounds).  (60) 

n  —  i 

Since  PF  =  wRT, 

W  =  wR  (Tl  "  T2)  (foot-pounds).  (61) 


n  —  i 


Equations  (60)  and  (61)  apply  to  any  gas  undergoing  expan- 
sion or  compression  according  to  PF71  =  a  constant.  In  the  case 
of  an  adiabatic  expansion  of  a  perfect  gas  n  =  y  (see  equation 

55)- 
Change    of    Internal    Energy   During    Adiabatic    Processes. 

Since  in  an  adiabatic  expansion  no  heat  is  conducted  to  or  away 
from  the  gas,  the  work  is  done  at  the  expense  of  the  internal 
energy  and,  therefore,  the  latter  decreases  by  an  amount  equiva- 
lent to  the  amount  of  work  performed.  This  loss  in  internal 
energy  is  readily  computed  by  equation  (6b)  or  (61).  The 
result  must  be  divided  by  778  in  order  to  be  in  B.t.u. 
During  an  adiabatic  compression  the  reverse  occurs,  i.e.,  there 


36  EXPANSION  AND   COMPRESSION  OF  GASES 

is  a  gain  in  internal  energy  and  the  same  formulas  apply,  the 
result  coming  out  negative,  because  work  has  been  done  on  the 
gas. 

Relation  between  Volume,  Pressure  and  Temperature  in 
Adiabatic  Expansion  of  a  Perfect  Gas.  Since  P,  V  and  T  vary 
during  an  adiabatic  expansion,  it  will  be  necessary  to  develop  for- 
mulas for  obtaining  these  various  quantities.  It  will  be  remem- 
bered that  equation  (25),  applies  to  perfect  gases  at  all  times. 
Therefore,  in  the  case  of  an  adiabatic  expansion  or  compression 
we  can  write  the  two  simultaneous  equations: 

PV      P2F 


22 


(B) 

By  means  of  these  two  equations  we  can  find  the  final  condi- 
tions of  pressure,  volume  and  temperature,  having  given  two 
initial  conditions  and  one  final  condition. 

For  instance,  having  given  FI,  F2  and  Ti}  to  find  T2,  divide  (A ) 
by  (B),  member  for  member.  Then 

F,  F2 


or  ft.Ti\j±)      •  (62) 

In  like  manner  the  following  formulas  can  be  obtained: 


ADIABATIC  EXPANSION  AND  COMPRESSION  37 

(65) 


P.- F,(^ 


It  should  be  noted  that  the  above  formulas  can  be  used  for 
any  expansion  of  a  perfect  gas  following  PVn  =  a  constant, 
provided  7  in  the  formulas  is  replaced  by  ». 

It  is  also  to  be  noted  that  these  equations  can  be  used  for 
any  system  of  units  so  long  as  the  same  system  of  units  is 
employed  throughout  an  equation. 

There  are  many  cases  of  expansions  which  are  neither  adiabatic 
nor  isothermal  and  which  are  not  straight  lines  on  P-V  diagrams. 
It  will  be  observed  from  the  equations  in  the  discussion  of  the 
internal  work  done  by  an  expanding  gas  and  for  the  change  of 
internal  energy,  that  if  in  the  general  equation  PVn  =  a  constant 
the  exponent  or  index  n  is  less  than  7,  the  work  done  is  greater 
than  the  loss  in  internal  energy.  In  other  words  for  such  a 
case,  the  expansion  lies  between  an  adiabatic  and  isothermal  and 
the  gas  must  be  taking  in  heat  as  it  expands.  On  the  other 
hand,  if  n  is  greater  than  7  the  work  done  is  less  than  the  loss 
of  internal  energy. 

Example.  Given  a  quantity  of  pure  air  in  a  cylinder  at  a  tem- 
perature of  60°  F.  (7\  =  460  +  60  =  5  20  degrees  absolute)  which 
is  suddenly  (adiabatically)  compressed  to  half  its  original  volume. 

y       2 
Then  —  =  -,  and  taking  7  as  1.405,  the  temperature  immediately 

after  compression  is  completed,  T2,  is  calculated  by  equation  (62) 
as  follows: 

/FA7"1  /2\1-406-1 

Tz  =  Ti  (y\      =  520  l^\         =  520  X  20-405  =  688°  absolute, 

or  /2  in  ordinary  Fahrenheit  is  688  —  460  or  228  degrees. 


38  EXPANSION  AND   COMPRESSION  OF   GASES 

The  work  done  in  adiabatic  compression  of  o±le  pound  of  this 
air  is  calculated  by  equation  (61)  : 

wR(T,-T,}  =  53.3  (520-688)  =  53.3  (-168)  =  _ 


7-1  1.405-1  0.405 

foot-pounds  per  pound  of  air  compressed.  TJie  negative  sign 
means  that  work  has  been  done  on  the  gas  or  the  gas  was  com- 
pressed. If  the  sign  had  been  positive  it  would  have  indicated 
an  expansion. 

As  the  result  of  this  compression  the  internal  energy  of  the  gas 

r\  sy     T  TO 

has  been  increased  by  -       -  B.t.u.,  but  if  .the  cylinder  is  a  con- 
778 

ductor  of  heat,  as  in  practice  it  always  is,  the  whole  of  this  heat 
will  become  dissipated  in  time  by  conduction  to  surrounding 
air  and  other  bodies,  and  the  internal  energy  will  gradually 
return  to  its  original  value  as  the  temperature  of  the  gas  comes 
back  to  the  initial  temperature  of  60°  F. 

PROBLEMS 

1.  Calculate  the  heat  required  to  produce  a  temperature  change  of  50°  F. 
in  three  pounds  of  air  at  constant  pressure. 

2.  How  many  foot-pounds  of  work  are  done  by  2  Ibs.  of  air  in  expand- 
ing to  double  its  volume  at  a  constant  temperature  of  100°  F.? 

3.  Three  pounds  of  air  are  to  be  compressed  from  a  volume  of  2  to  i  cu.  ft. 
at  a  constant  temperature  of  60°  F.    How  many  B.t.u.  of  heat  must  be 
rejected  from  the  air? 

4.  An  air  compressor  has  a  cylinder  volume  of  2  cu.  ft.    If  it  takes  air 
at  14.4  Ibs.  per  sq.  in.  absolute  and  70°  F.  and  compresses  it  isothermally  to 
100  Ibs.  per  sq.  in.  absolute,  find 

(a)  Pounds  of  air  in  cylinder  at  beginning  of  compression  stroke. 

(b)  The  final  volume  of  the  compressed  air. 

(c)  The  foot-pounds  of  work  done  upon  the  gas  during  compres- 

sion. 

(d)  The  B.t.u.  absorbed  by  the  air  in  increasing  the  internal  energy. 
(c)  The  B.t.u.  to  be  abstracted  from  the  cylinder. 

5.  A  cubic  foot  of  air  at  a  pressure  of  150  Ibs.  per  sq.  in.  gage  expands 
isothermally  until  its  pressure  is  50  Ibs  per  sq.  in.  gage.     Calculate  the 
work  done  during  this  expansion. 


PROBLEMS  39 

6.  Air  at  100  Ibs  per  sq.  in.  absolute  and  a  volume  of  2  cu.  ft.  expands 
along  an  n  =  i  curve  to  25  Ibs.  per  sq.  in.  absolute  pressure.    Find 

(a)  Work  done  by  the  expansion. 

(b)  Heat  to  be  supplied. 

7.  A  quantity  of  air  at  100  Ibs.  per  sq.  in.  absolute  pressure  has  a  tem- 
perature of  80°  F.    It  expands  isothermally  to  a  pressure  of  25  Ibs.  per 
sq.  in.  absolute  when  it  has  a  volume  of  4  cu.  ft.    Find  (i)  the  mass  of  air 
present,  (2)  Work  of  the  expansion  in  foot-pounds,  (3)  Heat  required  in 
B.t.u. 

8.  Air  at  100  Ibs.  per  sq.  in.  absolute  pressure  and  2  cu.  ft.  expands  to 
25  Ibs.  per  sq.  in.  absolute  adiabatically.    What  is  the  final  volume? 

9.  One  cubic  foot  of  air  at  60°  F.  and  a  pressure  of  15  Ibs.  per  sq.  in. 
absolute  is  compressed  without  loss  or  addition  of  heat  to  100  Ibs.  per  sq. 
in.  absolute  pressure.    Find  the  final  temperature  and  volume. 

10.  Two  pounds  of  air  are  expanded  from  a  temperature  of  300°  F.  to 
200°  F.  adiabatically.    How  many  foot-pounds  of  work  are  developed? 

11.  A  quantity  of  air  having  a  volume  of  i  cu.  ft.  at  60°  F.  under  a  pres- 
sure of  100  Ibs.  per  sq.  in.  absolute  is  expanded  to  5  cu.  ft.  adiabatically. 
Find  the  pounds  of  air  present,  the  final  temperature  of  the  air  and  the 
work  done  during  this  expansion. 

12.  Data   the  same  as  in  Problem  4  but  the  compression  is  to  be 
adiabatic.     Find 

(a)  The  final  volume  of  the  compressed  air. 

(b)  The  final  temperature  of  the  compressed  air. 

(c)  The  foot-pounds  of  work  required  to  compress  this  air. 

(d)  The  B.t.u.  absorbed  by  the  air  in  increasing  the  internal 

energy. 

(e)  The  B.t.u.  to  be  abstracted  from  the  gas. 

13.  A  pound  of  air  at  32°  F.  and  atmospheric  pressure  is  compressed  to 
4  atmospheres  (absolute).     What  will  be  the  final  volume  and  the  work 
of  compression  if  the  compression  is  (a)  isothermal,  (b)  adiabatic? 

14.  Plot  the  curve  PVn  =  C,  when  n  =  1.35.     Initial  pressure  is  460 
Ibs.  per  sq.  in.  gage,  initial  volume  0.5  cu.  ft.,  and  final  volume  8  cu.  ft. 

15.  Prove  that  the  work  of  an  adiabatic  expansion  can  be  expressed  by 
the  formula: 


CHAPTER  IV 
CYCLES  OF  HEAT  ENGINES  USING  GAS 

The  Heat  Engine  Cycle.  In  the  heat  engine,  the  working  sub- 
stance or  heat  medium  undergoes  changes  in  its  physical  proper- 
ties converting  heat  into  mechanical  work.  The  series  of  such 
changes  by  the  repetition  of  which  the  conversion  of  heat  into 
work  takes  place  forms  the  heat  engine  cycle.*  The  heat  engine 
cycle  usually  consists  of  four  events  which  are:  heating,  cooling, 
expansion,  and  compression. 

i.    THE  CARNOT  CYCLE 

Very  important  conclusions  regarding  theoretically  perfect 
heat  engines  are  to  be  drawn  from  the  consideration  of  the  action 
of  an  ideal  engine  in  which  the  working  substance  is  a  perfect 
gas  which  is  made  to  go  through  a  cycle  of  changes  involving 
both  isothermal  and  adiabatic  expansions  and  compressions. 
This  ideal  cycle  of  operations  was  invented  and  first  explained  in 
1824  by  Carnot,  a  French  engineer.  This  cycle  gave  the  first 
theoretical  basis  for  comparing  heat  engines  with  an  ideally 
perfect  engine.  The  ideal  Carnot  cycle  requires  an  engine, 
illustrated  in  Fig.  7,  which  consists  of  the  following  parts: 

(i)  A  piston  and  cylinder,  as  shown  in  Fig.  7,  composed  of 
perfectly  non-conducting  material,  except  the  cylinder-head  (left- 
hand  end  of  the  cylinder)  which  is  a  good  conductor  of  heat.  The 
space  in  the  cylinder  between  the  piston  and  the  cylinder  head  is 
occupied  by  the  working  substance,  which  can  be  assumed  to  be 
a  perfect  gas. 

*  A  thermodynamic  machine  performing  a  cycle  in  which  heat  is  changed  into 
work  is  called  a  heat  engine,  and  one  performing  a  cycle  in  which  heat  is  trans- 
ferred from  a  medium  at  a  low  temperature  to  one  at  a  higher  temperature  is 
called  a  refrigerating  machine. 

40 


THE  CARNOT  CYCLE 


(2)  A  hot  body  H  of  unlimited  heat  capacity,  always  kept  at  a 
temperature  T\. 

(3)  A  perfectly  non-conducting  cover  N. 

(4)  A  refrigerating  or  cold  body  R  of  unlimited  heat  receiving 
capacity,  which  is  kept  at  a  constant  temperature  T2  (lower 
than  Ti). 


FIG.  7.  —  Apparatus  and  Diagram  Illustrating  a  Reversible  (Carnot)  Cycle. 

It  is  arranged  that  H,  N  or  R  can  be  applied,  as  required,  to 
the  cylinder  head.  Assume  that  there  is  a  charge  of  one  pound 
of  gas  in  the  cylinder  between  the  piston  and  the  cylinder  head, 
which  at  the  beginning  of  the  cycle,  with  the  piston  in  the  posi- 
tion shown,  is  at  the  temperature  TI,  has  a  volume  Va,  and  has  a 
pressure  Pa.  The  subscripts  attached  to  the  letters  V  and  P 
refer  to  points  on  the  pressure-volume  diagram  shown  in  Fig.  7. 


42  CYCLES  OF  HEAT  ENGINES   USING  GAS 

This  diagram  shows,  by  curves  connecting  the  points  a,  b,  c  and 
d,  the  four  steps  in  the  cycle. 

The  operation  of  this  cycle  will  be  described  in  four  parts  as 
follows: 

(1)  Apply  the  hot  body  or  heater  H  to  the  cylinder  head  at 
the  left-hand  side  of  the  figure.    The  addition  of  heat  to  the  gas 
will  cause  it  to  expand  isothermally  along  the  curve  ab,  because 
the  temperature  will  be  maintained  constant  during  the  process 
at  TI.    The  pressure  drops  slightly  to  P*  when  the  volume  be- 
comes V».     During  this  expansion  external  work  has  been  done 
in  advancing  the  piston  and  the  heat  equivalent  of  this  work  has 
been  obtained  from  the  hot  body  H. 

(2)  Take  away  the  hot  body-H  and  at  the  same  time  attach 
to  the  cylinder  head  the  non-conducting  cover  N.    During  this 
time  the  piston  has  continued  to  advance  toward  the  right, 
doing  work  without  receiving  any  heat  from  an  external  source, 
so  that  the  expansion  of  the  gas  in  this  step  has  been  done  at 
the  expense  of  the  stock  of  internal  energy  in  the  gas  along  the 
adiabatic  curve  be.    The  temperature  has  continued  to  drop  in 
proportion  to  the  loss  of  heat  to  the  value  T2.     Pressure  is  then 
Pc  and  the  volume  is  Vc. 

(3)  Take  away  the  non-conductor  N  and  apply  the  refriger- 
ator R.    Then  force  the  piston  back  into  the  cylinder.    The 
gas  will  be  compressed  isothermally  at  the  temperature   T2. 
In  this  compression,  work  is  being  done  on  the  gas,  and  heat  is 
developed,  but  all  of  it  goes  into  the  refrigerator  R,  in  which  the 
temperature  is  always  maintained  constant  at  TV    This  com-, 
pression  is  continued  up  to  a  point  d  in  the  diagram,  so  selected 
that  a  further  compression    (adiabatic)   in   the  next   (fourth) 
stage  will  cause  the  volume,  pressure  and  temperature  to  reach 
their  initial  values  as  at  the  beginning  of  the  cycle.* 

(4)  Take  away  the  refrigerator  R  and  apply  the  non-conduct- 
ing cover  N.     Then  continue  the  compression  of  the  gas  without 

*  Briefly  the  third  stage  of  the  cycle  must  be  stopped  when  a  point  d  is  reached, 
so  located  that  an  adiabatic  curve  (PV7  -  constant)  drawn  from  it  will  pass 
through  the  "initial"  point  a. 


THE   CARNOT   CYCLE  43 

the  addition  of  any  heat.  It  will  be  the  adiabatic  curve  da. 
The  pressure  and  the  temperature  will  rise  and,  if  the  point  d  has 
been  properly  selected,  when  the  pressure  has  been  brought  back 
to  its  initial  value  Pa  the  temperature  will  also  have  risen  to  its 
initial  value  T\.  The  cycle  is  thus  finished  and  the  gas  is  ready 
for  a  repetition  of  the  same  series  of  processes  comprising  the 
cycle. 

To  define  the  Carnot  cycle  completely  we  must  determine 
how  to  locate  algebraically  the  proper  place  to  stop  the  third 
step  (the  location  of  d).  During  the  second  step  (adiabatic  ex- 
pansion from  b  to  c)  by  applying  equation  (62),  the  following 
temperature  and  volume  relations  exist: 


also  for  the  adiabatic  compression  the  fourth  step  can  be  simi- 
lary  stated, 


Hence, 

(T  -RF' 

Simplifying  and  transposing,  we  have 

V*_      Vc 

Va    "    Vd' 

— *  is  the  ratio  of  expansion  r  for  the  isothermal  expansion  in 

V  d 

the  first  step  of  the  cycle.     This  has  been  shown  (Equation  68)  to 

Vc 
be  equal  to  -e-  in  the  isothermal  compression  in  the  third  step  in 

order  that  the  adiabatic  compression  occurring  in  the  fourth 
step  shall  complete  the  cycle. 

A  summary  of  the  heat  changes  to  and  from  the  working 
gas  (per  pound)  in  the  four  steps  of  the  Carnot  cycle  is  as  fol- 
lows: 


44  CYCLES  OF  HEAT  ENGINES   USING  GAS 

(ab).    Heat  taken  in  from  hot  body  (by  equation  (45),  in 
foot-pounds)  is: 

RT^oge^-  (69) 

V  a 

(be).     No  heat  taken  in  or  rejected. 

(cd).     Heat  rejected  to  refrigerator  (by  equation  (45),  in 
foot-pounds)  is: 

*r,iog,£=  -#r2ioge£.*  (7o) 

V  c  V  d 

(da).     No  heat  taken  in  or  rejected. 

Hence,  the  net  amount  of  work  done,  W,  by  the  gas  in  this 
cycle,  being  the  mechanical  equivalent  (foot-pounds)  of  the 
excess  of  heat  taken  in  over  that  rejected,  is  the  algebraic  sum 

of  (69)  and  (70): 

W  =  R(TI  ioge  £  -  r2  ioge  ^)  -  *  (zv  -  zv)  log,  £•     (71) 

The  thermal  or  heat  efficiency  of  a  cycle  is  defined  as  the  ratio  of 

Heat  equivalent  of  work  done 
Heat  taken  in 

The  heat  equivalent  of  work  done  is,  by  equation  (71), 


and  the  heat  taken  in  is,  by  equation  (69), 

tfTilog.^- 

V  a 

The  ratio  above  representing  the  efficiency  E  is: 


REVERSIBLE   CYCLES  45 

The  efficiency  of  the  Carnot  cycle  as  expressed  by  equation  (72) 
is  the  maximum  possible  theoretical  efficiency  which  may  be 
obtained  with  any  heat  engine  working  between  the  temperature 
limits  TI  and  T2.  This  equation  can  be  used  as  a  standard  for  the 
comparison  of  the  efficiencies  of  actual  heat  engines  working 
between  two  temperature  limits. 

Reversible  Cycles.  A  heat  engine  which  is  capable  of  dis- 
charging to  the  "source  of  heat"  when  running  in  the  reverse 
direction  from  that  of  its  normal  cycle  the  same  quantity  of 
heat  that  it  would  take  from  this  source  when  it  is  running 
direct  and  doing  work  is  said  to  operate  with  its  cycle  reversed, 
or,  in  other  words,  the  engine  is  reversible.  A  reversible  heat 
engine  then  is  one  which,  if  made  to  follow  its  indicator  diagram 
in  the  reverse  direction,  will  require  the  same  horse  power  to  drive 
it  as  a  refrigerating  machine  as  the  engine  will  deliver  when 
running  direct,  assuming  that  the  quantity  of  heat  used  is  the 
same  in  the  two  cases.  An  engine  following  Carnot's  cycle  is, 
for  example,  a  reversible  engine.  The  thermodynamic  idea  of 
reversibility  in  engines  is  of  very  great  value  because  no  heat 
engine  can  be  more  efficient  than  a  reversible  engine  when  both 
work  between  the  same  limits  of  temperature;  that  is,  when 
both  engines  take  in  the  same  amount  of  heat  at  the  same  higher 
temperature  and  reject  the  same  amount  at  the  same  lower 
temperature. 

It  was  first  proved  conclusively  by  Carnot  that  no  other  heat 
engine  can  be  more  efficient  than  a  reversible  engine  when  both 
work  between  the  same  temperature  limits.  To  illustrate  this 
principle,  assume  that  there  are  two  engines  A  and  B.  Of  these 
let  us  say  A  is  reversible  and  B  is  not.  In  their  operation  both 
take  heat  from  a  hot  body  or  heater  H  and  reject  heat  to  a  refrig- 
erator or  cold  body  R.  Let  QH  be  the  quantity  of  heat  which 
the  reversible  engine  A  takes  in  from  the  hot  body  H  for  each 
unit  of  work  performed,  and  let  QR  be  the  quantity  of  heat  per 
unit  of  work  which  it  discharges  to  the  refrigerator  R. 

For  the  purpose  of  this  discussion,  assume  that  the  non- 
reversible  engine  B  is  more  efficient  than  the  reversible  engine  A. 


46  CYCLES   OF  HEAT  ENGINES   USING  GAS 

Under  these  circumstances  it  is  obvious  that  the  engine  B  will  take 
in  less  heat  than  A  and  it  will  reject  correspondingly  less  heat 
to  R  per  unit  of  work  performed.  The  heat  taken  in  by  the 
non-reversible  engine  B  from  the  hot  body  H  we  shall  designate 
then  by  a  quantity  less  than  QH,  or  QH  —  X,  and  the  heat  rejected 
by  B  to  the  refrigerator  R  by  QR  —  X.  Now  if  the  non-reversible 
engine  B  is  working  direct  (when  converting  heat  into  work) 
and  is  made  to  drive  the  reversible  engine  A  according  to  its 
reverse  cycle  (when  converting  work  into  heat),  then  for  every 
unit  of  work  done  by  the  engine  B  in  driving  the  reversible 
engine  A,  the  quantity  of  heat  mentioned  above,  that  is,  QH  —  X, 
would  be  taken  from  the  hot  body  H  by  the  non-reversible 
engine  B  and,  similarly,  the  quantity  of  heat  represented  by  QH 
would  be  returned  to  the  hot  body  H  by  the  reverse  action  of 
the  cycle  of  operations  performed  by  A.  This  follows  because 
the  engine  A  is  reversible  and  it  returns,  therefore,  to  H,  when 
operating  on  the  reverse  cycle,  the  same  amount  of  heat  as  it 
would  take  in  from  H  when  working  on  its  direct  cycle.  By 
this  arrangement  the  hot  body  H  would  be  continually  receiving 
heat,  in  the  amount  represented  by  X  for  each  unit  of  work 
performed.  At  the  same  time  the  non-reversible  engine  B  dis- 
charges to  the  refrigerator  R  a  quantity  of  heat  represented  by 
QR  —  X,  while  the  reversible  engine  A  removes  from  the  refriger- 
ator R  a  quantity  represented  by  QR.  As  a  result  of  this  last 
operation  the  cold  body  will  be  losing  continually  per  unit  of  work 
performed  a  quantity  of  heat  equal  to  X.  The  combined  per- 
formances of  the  two  engines,  one  working  direct  as  a  normal  heat 
engine  and  the  other,  according  to  its  reverse  cycle,  as  a  com- 
pressor or  what  might  be  called  a  "heat  pump,"  gives  a  constant 
removal  of  heat  from  the  refrigerator  R  to  the  hot  body  H, 
and  as  a  result  a  degree  of  infinite  coldness  must  be  finally  pro- 
duced in  the  refrigerator. 

If  we  assume  that  there  is  no  mechanical  friction,  this  com- 
bined machine,  consisting  of  a  normal  heat  engine  and  com- 
pressor, will  require  no  power  from  outside  the  system.  For 
this  reason  the  assumption  that  the  non-reversible  engine  B 


REVERSIBLE   CYCLES  47 

can  be  more  efficient  than  the  reversible  engine  A  has  brought 
us  to  a  result  which  is  impossible  from  the  standpoint  of  expe- 
rience as  embodied  hi  the  statement  of  the  "  Second  Law  of 
Thermodynamics  ";  that  is,  it  is  impossible  to  have  a  self-acting 
engine  capable  of  transferring  heat,  infinite  in  quantity,  from  a 
cold  body  to  a  hot  body.  We  should,  therefore,  conclude  that 
no  non-reversible  engine,  as  B  for  example,  can  be  more  efficient 
than  a  reversible  engine  A  when  both  engines  operate  between 
the  same  temperature  limits.  More  briefly,  when  the  source  of 
heat  and  the  cold  receiver  are  the  same  for  both  a  reversible  heat 
engine  and  any  other  engine,  then  the  reversible  engine  must 
have  a  higher  possible  efficiency;  and  if  both  engines  are  reversible 
it  follows  that  neither  can  be  more  efficient  than  the  other. 

A  reversible  engine  is  perfect  from  the  viewpoint  of  efficiency; 
that  is,  its  efficiency  is  the  best  obtainable.  No  other  engine 
than  a  reversible  engine  which  takes  in  and  discharges  heat  at 
identical  temperatures  will  transform  into  work  a  greater  part  of 
the  heat  which  it  takes  in.  Finally,  it  should  be  stated  as  regards 
this  efficiency  that  the  nature  of  the  substance  being  expanded  or 
compressed  has  absolutely  no  relation  to  the  thermal  efficiency 
as  outlined  above. 

If  an  engine  operating  on  Carnot's  cycle  is  reversed  in  its  action, 
so  that  the  same  indicator  diagram  shown  in  Fig.  7  would  be 
traced  in  the  opposite  direction,  the  reversed  cycle,  when  begin- 
ning as  before  at  point  a  with  a  perfect  gas  at  the  temperature 
Zi,  will  consist  of  the  following  stages: 

(1)  When  the  non-conductor  N  is  applied  and  the  piston  is 
advanced  toward  the  right  by  the  source  of  power  performing 
the  reversed  cycle,  the  gas  will  expand,  tracing  the  adiabatic 
curve  ad,  with  constant  lowering  of  temperature  which  at  the 
point  d  will  be  T2. 

(2)  When  the  non-conductor  N  is  now  removed,  the  refrig- 
erator R  is  applied,  and  the  piston  continues  on  its  outward  stroke. 
The  gas  will  expand  isothermally  at  the  constant  temperature  T2, 
tracing  the  curve  dc.     During  this  stage  the  gas  is  taking  heat 
from  the  refrigerator  R. 


48  CYCLES   OF  HEAT  ENGINES   USING  GAS 

(3)  When  the  refrigerator  R  is  removed  and  the  non-conductor 
N  is  again  applied,  which  will  be  on  the  back  stroke  of  the  engine, 
the  gas  will  be  compressed,  and  on  the  indicator  diagram  another 
adiabatic  curve  cb  will  be  traced.    At  the  point  b  the  temper- 
ature will  be  obviously  T\. 

(4)  When  the  non-conductor  N  is  removed  and  the  hot  body 
H  is  again  applied,  with  the  compression  continuing  along  the 
isothermal  curve  ba,  heat  will  be  discharged  to  the  hot  body  H, 
while  the  temperature  is  maintained  constant  at  7\. 

The  cycle  has  now  been  traced  in  a  reverse  direction  from  the 
beginning  back  to  the  starting  point  at  a,  and  is  now  complete. 
During  this  process  an  amount  of  work  represented  by  the  area 
of  the  indicator  diagram,  equivalent  in  foot-pounds  to 

R  loge  ^  (Ti  -  T2)  (see  equation  71), 

V  a 

has  been  converted  into  heat.     First,  heat  was  taken  from  the 
refrigerator  R,  represented  in  amount  by 


and  second,  heat  was  rejected  to  the  hot  body  H  in  the  amount 


or     - 


As  in  direct  operation  of  Carnot's  cycle  no  heat  is  given  or  lost 
in  the  first  and  third  stages  outlined  above.  The  algebraic  sum 

of  these  two  quantities,  remembering  that  —  ^  =  —  -,  gives  the 

V  a  V  d 

net  amount  of  work  done,  W,  on  the  gas,  and,  therefore,  the 
net  amount  of  heat  (foot-pound  units)  transferred  from  the  cold 
body  R  to  the  hot  body  H  or, 

W  =  RT2  loge  ^  -  RT,  log,  ^=   -R  log,  -p  (Ti  -  T2).    (73) 


HOT-AIR   ENGINE   CYCLES  49 

Since  the  result  is  the  same  as  given  by  equation  (72),  although 
opposite  in  sign  on  account  of  being  work  of  compression,  it 
will  be  observed  that  in  the  reverse,  cycle  the  same  amount 
of  heat  is  given  to  the  hot  body  H  as  was  taken  from  it  in  the 
direct  operation  of  the  same  cycle,  and  that  the  same  amount 
of  heat  is  now  taken  from  the  refrigerator  R  as  was  in  the  other 
case  ,given  to  it. 

2.     HOT-AIR  ENGINE   CYCLES 

The  hot-air  engine  is  a  non-explosive  type  of  external  combus- 
tion heat  engine,  the  working  substance  being  atmospheric  air 
which  undergoes  no  change  in  its  physical  state.  This  type  of 
prime-mover  was  invented  about  100  years  ago.  It  is  little  used 
on  account  of  the  difficulty  in  transmitting  heat  through  the 
metallic  walls  to  the  dry  gas  air.  Then  the  hot-air  engine  is 
very  bulky  in  proportion  to  its  power.  There  are  also  difficul- 
ties in  carrying  out  practically  the  theoretical  cycles  of  operation, 
on  account  of  the  rapid  deterioration  of  the  heat-conducting  sur- 
faces. The  mechanical  efficiency  of  the  hot-air  engine  is  also  low. 

The  advantages  of  the  hot-air  engine  are  ease  of  operation  and 
safety.  Also  its  speed  being  low,  on  account  of  the  rate  of  heat 
transmission,  it  is  well  suited  for  the  driving  of  small  pumps  and 
for  other  domestic  uses  where  small  powers  are  required  and  also 
where  the  fuel-economy  is  a  matter  of  minor  importance. 

Hot-air  engine  cycles  are  divided  into  two  groups:  - 

Group  I.  External-combustion  hot-air  engines  with  a  closed 
cycle  and  constant-volume  temperature  changes. 

Group  II.  External-combustion  hot-air  engines  with  an  open 
cycle  and  constant-pressure  temperature  changes. 

In  the  hot-air  engines  an  attempt  was  made  to  put  in  practice 
a  cycle  of  the  ideal  Carnot  type  with  the  addition  of  a  regenera- 
tor. The  regenerator  was  a  storage-battery  of  heat,  its  function 
being  to  absorb,  store  and  return  heat  rapidly,  replacing  the 
adiabatic  expansion  curves  of  the  Carnot  cycle  by  lines  of  con- 
stant volume  in  group  I  and  by  lines  of  constant  pressure  in 


CYCLES   OF  HEAT  ENGINES   USING   GAS 


group  II.  The  regenerator  consists  of  a  chamber  containing 
strips  of  metal,  coils  of  wire,  or  any  other  heat-absorbing  material 
arranged  in  such  a  manner  as  to  present  a  very  large  surface  to 
the  air  passing  through  it. 

The  Stirling  Engine  Cycle.  The  Stirling  engine  belongs  to 
group  I,  the  regenerator  being  so  arranged  that  the  pressure 
drops  with  the  temperature  at  such  a  rate  as  to  keep  the  vol- 
ume constant.  It  is  an  external-combustion  engine  and  its 


Volume 
FIG.  8.  —  Stirling  Engine  Cycle. 

cycle  of  operation  is  closed,  the  same  air  is  used  over  and 
over  again,  any  loss  by  leakage  being  supplied  by  a  small 
force-pump.  The  engine  consists,  essentially,  of  a  pair  of  dis- 
placer  cylinders,  a  double-acting  working  cylinder,  a  regenera- 
tor and  a  refrigerator.  In  the  earlier  forms  the  displacer  cyl- 
inder had  a  double  wall,  the  regenerator  and  refrigerator  being 
placed  in  the  annular  space  surrounding  the  displacer  cylinder. 
In  the  latter  forms  the  regenerator  and  refrigerator  are  arranged 
separately  from  the  displacer  cylinder  but  in  direct  communi- 
cation with  its  top  and  bottom.  A  plunger  works  in  the  dis- 
placer cylinder,  this  plunger  being  filled  with  a  non-conducting 
material  such  as  brick-dust.  The  engine  derives  its  heat  by 


STIRLING  ENGINE   CYCLE  51 

conduction  from  a  furnace  which  is  placed  beneath  the  displac- 
ing cylinder. 

In  the  cycle  represented  by  Fig.  8,  the  action  of  the  engine 
is  carried  out  as  follows: 

1.  The  substance,  air,  having  a  volume  Va,  a  pressure  Pa  and 
a  temperature  Ta  receives  heat  at  constant  volume  by  passing 
through  the  regenerator,  and  receives  also  heat  from  the  furnace. 
As  a  result  of  the  heat  addition,  its  pressure  is  increased  to  Pbj  its 
temperature  to  Tb,  this  process  being  represented  by  the  constant 
volume  line  ab,  Fig.  8.   During  this  event  the  plunger  is  moving  up. 

2.  The  pressure  under  the  working  piston  increases  as  a  result 
of  the  addition  of  heat  and  the  expansion  of  the  air  follows, 
producing  the  working  stroke.     This  expansion  is  represented  by 
be  and  is  an  isothermal,  the  air  receiving  heat  from  the  furnace 
plates  and  the  temperature  remaining  TV 

3.  The  plunger  descends,  displacing  the  hot  air  and  forcing  the 
same  through  the  regenerator  and  refrigerator.     As  a  result  of 
this,  the  temperature  drops  from  Tc  to  Td  and  the  pressure  is 
decreased  from  Pc  to  Pd,  the  volume  remaining  constant  as  shown 
by  the  line  cd. 

4.  Due  to  this  cooling  effect  the  working  piston  descends  com- 
pressing the  air.     This  compression,  represented  by  da,  is  iso- 
thermal, the  air  being  cooled  during  the  process  by  the  regenera- 
tor and  refrigerator  to  remain  at  the  temperature  TV 

Referring  to  Fig.  8  of  the  Stirling  cycle,  the  useful  work  W 
is  in  foot-pounds, 

(74) 


The  mean  effective  pressure  is  in  pounds  per  square  foot, 

W 


The  horse  power  developed  can  be  represented  by 
Rp   _WxN 

33,000 
where  N  —  revolutions  (cycles)  per  minute. 


W  v  N 

H.P.  =  w  X      .  (76) 

33,000 


CYCLES  OF  HEAT  ENGINES  USING  GAS 


Since  the  expansion  and  compression  processes  are  isothermal, 
the  efficiency  of  the  Stirling  cycle  is: 


E  = 


"F, 


>eV0 


(77) 


Since   Va  =  Vb  and   Vc  =  Vd,  also  RTi  =  P,V,  and   RT,  = 
aVdj  equation  (77)  becomes, 


E  = 


(78) 


The  Ericsson  Engine  Cycle.  This  engine  belongs  to  group  II, 
the  regenerator  changing  the  temperature  at  constant  pressure. 
This  engine  consists  of  five  parts :  a  compressing  pump,  a  receiver, 
a  regenerator,  a  refrigerator  and  a  working  cylinder.  The  ideal 


Volume 
FIG.  9.  —  The  Ericsson  Engine  Cycle. 

cycle  for  this  engine  is  represented  by  Fig.  9,  the  order  of  events 
being  as  follows : 

Atmospheric  air  is  drawn  into  the  compressing  pump  as  shown 
by  d'd.  This  air  is  compressed  isothermally  to  0,  during  the 
return  stroke  of  the  pump;  it  is  then  forced  into  a  receiver  as 
shown  by  aaf .  Thus  d'daa!  represents  the  pump  cycle. 

As  the  working  cylinder  begins  its  forward  or  up  stroke,  the 


THE  LENOIR  ENGINE  53 

compressed  air  is  admitted  into  the  working  cylinder  at  constant 
pressure.  As  this  admission  from  a  to  b  is  through  the  regenera- 
tor, the  entering  air  takes  up  heat  increasing  in  temperature  from 
T2  to  T\.  As  soon  as  the  supply  of  compressed  air  is  cut  off, 
the  air  expands  isothermally  along  be  to  atmospheric  pressure, 
while  heat  is  being  supplied  by  a  furnace  at  the  bottom  of  the 
working  cylinder.  During  the  return  stroke  of  the  working  pis- 
ton, the  air  is  discharged  at  constant  pressure  through  the  regen- 
erator, giving  up  its  heat  and  is  cooled  to  the  temperature  TV 
In  Fig.  9,  a' bed'  is  the  cycle  of  the  working  cylinder,  the  net  work 
being  represented  by  abed. 

The  Ericsson  cycle  has  the  same  efficiency  as  the  Stirling  cycle, 
the  regenerator  process  being  carried  out  at  constant  pressure 
instead  of  at  constant  volume. 

3.     INTERNAL-COMBUSTION  ENGINE   CYCLES 

The  internal-combustion  engine  utilizes  as  its  working  sub- 
stance a  mixture  of  air  and  gas  or  air  and  petroleum  vapor. 
Combustion  of  the  mixture  takes  place  inside  the  engine  cylinder, 
or  in  a  communicating  vessel,  and  the  heat  generated  is  converted 
into  work.  As  the  specific  heat  and  the  specific  volume  of  the 
mixture  do  not  differ  much  from  that  of  air,  the  cyclical  analy- 
sis and  the  theory  of  the  internal-combustion  engine  are  developed 
on  the  assumption  that  the  working  substance  is  air.  The 
internal-combustion  engine  cycles  can  be  divided  into  the  follow- 
ing groups: 

Group  I.     Engines  without  compression. 

Group  II.     Compression  engines. 

Group  I.    Internal-Combustion  Engines  Without  Compression 

The  Lenoir  Engine.  An  engine  of  this  group,  invented  by 
Pierre  Lenoir  in  1860,  was  the  first  successful  practical  gas 
engine.  The  action  of  this  type  of  engine  is  as  follows: 

As  the  piston  leaves  the  dead  center  it  draws  into  the  cylinder 
a  charge  of  gas  and  air  which  is  proportioned  at  the  admission 


54 


CYCLES  OF  HEAT  ENGINES  USING  GAS 


valve  to  form  an  explosive  mixture.  Admission  is  cut  off  some- 
what before  half-stroke  and  this  is  followed  by  the  ignition  of  the 
mixture  by  means  of  an  electric  spark.  The  explosion  produces 
a  rapid  rise  in  pressure  above  atmospheric,  thus  forcing' the  piston 
to  the  end  of  the  stroke.  The  exhaust  valve  opens  near  the  end 
of  the  stroke  and  the  burnt  products  are  expelled  during  the  return 
stroke  of  the  piston.  A  fly-wheel  carries  the  piston  over  during 
the  exhaust  stroke  as  well  as  also  during  the  suction  stroke. 
The  above  series  of  operations  takes  place  at  each  end  of  the 
piston,  producing  two  impulses  for  each  revolution. 

Fig.  10  represents  the  cycle  of  operations  of  the  Lenoir  engine 


Volume 
FIG.  10.  —  Ideal  Lenoir  Cycle. 

on  P-V  coordinates.  The  drawing  in  of  the  mixture  of  gas  and 
air  at  atmospheric  pressure  is  represented  by  ab.  Ignition  takes 
place  at  b  and  is  followed  by  the  constant  volume  combustion 
line  be.  The  working  stroke,  an  adiabatic  expansion,  takes  place 
to  atmospheric  pressure,  as  shown  by  cd,  and  the  products  of 
combustion  are  rejected  during  the  return  stroke  of  the  piston  da. 
Calling  P,  V,  T  the  absolute  pressure  in  pounds  per  square  foot, 
the  specific  volume  in  cubic  feet  per  pound  of  mixture,  and  the 
absolute  temperature  in  degrees  Fahrenheit  respectively;  also 
using  subscripts  b,  c,  d  to  designate  the  points  at  the  corners  bed 
of  Fig.  10,  the  following  expressions  will  be  obtained: 


THE  LENOIR  ENGINE  55 

If  the  heat  developed  by  the  complete  combustion  of  the 
mixture  be  designated  by  Qi,  then 

&  =  wC.  (Tc  -  TV).  (79) 

Since  the  combustion  of  the  mixture  takes  place  at  constant 
volume, 

Vc=  Vb 

Tc  =  Tb  +  %  (80) 

Co 

For  unit  weight, 

P*-P£-  <«i) 

lb 

The  expansion  cd  being  adiabatic  (n  =  7), 


-  '-®f 


If  ft  is  the  heat  rejected  by  the  engine  after  performing  its 
cycle  of  operations, 

ft  =  wCP  (Td  -  Tb).  (84) 

The  useful  work  available  is: 

W  =  J(Ql-Q2).  (85) 

The  mean  effective  pressure  is  the  average  unbalanced  pressure 
on  the  piston  of  the  engine  in  pounds  per  unit  area,  and  using 
symbols  and  units  as  on  page  51, 

M.E.P.  =  ir^-ir-  (86) 

"  d  V  6 

The  horse  power  developed  would  be  found  by  equation 

H.P.  =  W  X  N,  (87) 

33,ooo 

where  N  designates  the  number  of  explosions  per  minute. 
The  cycle  efficiency  for  the  conditions  of  the  problem  is: 

E.-I--t-yT  (88) 


CYCLES  OF  HEAT  ENGINES  USING  GAS 


Group  II.     Compression  Engines 

The  first  compression  engine  was  patented  as  early  as  1799  by 
Philip  Lebon.  The  thermodynamic  advantages  of  compression 
before  ignition  will  be  evident  from  the  demonstrations  which 
follow.  A  mechanical  advantage  due  to  compression  is  the  re- 
duced shock  due  to  the  explosion  and  the  resulting  improved 
balance  of  parts. 

i.  The  Otto  Cycle.  The  Otto  cycle,  embodying  the  principles 
first  proposed  by  Beau  de  Rochas  in  1862,  resulted  in  the  most 
successful  internal-combustion  engine  of  to-day.  This  cycle  was 
adopted  by  Dr.  Otto  in  1876,  the  operations  being  carried  out  in 
four  consecutive  equal  strokes  of  the  piston,  requiring  two  com- 
plete revolutions  of  the  engine  crank-shaft.  The  ideal  diagram 


Volume 
FIG.  ii.  — Otto  Cycle. 

for  the  Otto  cycle  takes  the  form  shown  by  Fig.  n,  the  operation 
being  as  follows: 

A  mixture  of  gas  and  air  is  drawn  in  during  the  complete  for- 
ward stroke  of  the  piston,  as  shown  by  a' a.  The  return  of  the 
piston  compresses  the  mixture  along  the  adiabatic  curve  ab. 
Explosion  of  the  compressed  charge  takes  place  at  b,  with  the 
consequent  combustion  at  constant  volume  to  c.  cd  is  the  adia- 


THE  BRAYTON  CYCLE  57 

batic  expansion  producing  the  second  forward  stroke.  The  ex- 
haust valve  opens  at  d,  cooling  the  gases  to  the  exhaust  pressure  a, 
and  rejecting  them  to  the  atmosphere. 

The  heat  added  during  the  combustion  from  b  to  c  is 

ft  =  wC,  (Tc  -  T>).  (89) 

The  heat  rejected  from  d  to  a  is 

Q2  =  wCv  (Td  -  Ta).  (90) 

The  cycle  efficiency  is 
c,  _  ft  —  ft  _  wCv  (Tc  —  Td)  —wCt  (Td  —  Ta)  _^_  Td  —  Ta         ,* 


Ql  wCv(Tc-Th)  Tc-T 

Since  the  expansion  and  compression  are  adiabatic  the  follow- 
ing relations  will  hold: 

TaVan~l  =  r^F,"-1,     and     TcVcn~l  =  TAVf~l. 
Since         Vc  =  Vb  and  F«  =  Vd, 

Td_Tc.  Tg-Ta  _  Ta       i/ 

T~  T'  Tc-T    ~  T 


Smce 


"'1  (p*\—     (    \ 

=I-n'   (92) 


Equation  (92)  shows  that  the  efficiency  of  the  Otto  cycle  de- 
pends on  the  amount  of  compression  before  explosion. 

2.  The  Brayton  Cycle.  This  cycle  is  often  called  the  Joule 
cycle  after  its  inventor,  or  the  Brayton  cycle  after  George  B. 
Brayton,  who  in  1872  designed  an  engine  with  gradual  constant 
pressure  combustion.  In  this  engine  a  mixture  of  gas  and  air 
is  first  compressed  in  a  separate  pump  and  forced  into  a  receiver. 
On  the  way  from  the  receiver  to  the  engine  cylinder  the  mixture  is 
ignited  by  a  gas  jet  which  burns  steadily  without  sudden  explo- 
sion, producing  temperature  and  volume  changes  at  constant 
pressure.  After  expansion  the  piston  drives  out  the  products 
of  combustion  at  atmospheric  pressure. 

The  cycle  of  operations  for  the  Brayton  engine  is  represented 


58  CYCLES  OF  HEAT  ENGINES  USING  GAS 

by  Fig.  12.  a' a  represents  the  supply  of  the  combustible  mixture 
to  the  pump  where  it  is  compressed  adiabatically  to  b  and  forced 
into  a  receiver,  be  represents  the  burning  of  the  compressed 


l' 


a' 


Volume 
FIG.  12.  —  Brayton  Cycle. 

mixture  at  constant  pressure.  As  the  mixture  enters  the  working 
cylinder,  it  expands  adiabatically  along  cd.  This  is  followed  by 
the  rejection  of  the  burnt  gases  along  the  atmospheric  line  da. 

The  heat  added  during  the  constant  pressure  combustion  from 
b  to  c  is 

Qi  =  wCP  (fc  -  Tb).  (93) 

The  heat  rejected  from  d  to  a  is 

ft  =  wCP  (Td  -  Ta).  (94) 

The  cycle  efficiency  is 


p  = 


Qi  -  Q* 


c-T>)  -wC 


wCP  (Tc  - 


Td-Ta 
Tc-  T 


,    . 


Since  the  expansion  and  compression  phases  of  the  cycle  are 
adiabatic  and  Pc  =  Pb\  Pa  =  Pa,  therefore, 


and 


Tt-Ta 


-1 


(96) 


Equation  (96)  is  the  same  as  equation  (92)  and  shows  that  the 
cycle  efficiencies  of  the  Otto  and  Brayton  cycles  are  the  same, 


THE  DIESEL  CYCLE 


59 


under  the  same  initial  conditions,  and  depend  on  the  amount  of 
compression  of  the  charge  before  explosion. 

3.  The  Diesel  Cycle.  This  cycle  is  carried  out  in  four  strokes 
of  equal  length,  as  in  the  case  of  the  Otto  cycle.  Atmospheric 
air  is  drawn  in  during  the  complete  forward  stroke  of  the  piston 
as  shown  by  la  (Fig.  13).  The  return  of  the  piston  compresses 


FIG.  13.  —  Diesel  Engine  Cycle. 

the  air  adiabatically  to  £,  a  pressure  of  about  500  pounds  per 
square  inch.  At  the  end  of  the  compression  stroke  a  charge  of 
the  liquid  fuel  is  injected  in  a  finely  divided  form  by  an  auxiliary 
pump  or  compressor  and  burns  nearly  at  constant  pressure  when 
it  conies  in  contact  with  the  highly  compressed  air.  The  supply 
and  combustion  of  the  fuel,  as  represented  by  the  combustion 
line  be,  is  cut  off  at  one-tenth  to  one-sixth  the  working  stroke, 
depending  upon  the  load.  Expansion  takes  place  during  the 
balance  of  the  stroke  as  shown  by  the  adiabatic  expansion  line  cd. 
Release  occurs  at  d  with  the  consequent  drop  in  pressure,,  and  the 
burnt  gases  are  rejected  during  the  fourth  stroke  of  the  cycle. 
The  heat  added  (ft)  during  the  combustion  of  the  fuel  is: 

ft  =  wcp  (TC  -  r>).  (97) 

The  heat  rejected  from  d  to  a  is 

ft  =  wCD  (Td  -  r«).  (98) 

The  cycle  efficiency  is  then, 


60  CYCLES  OF  HEAT  ENGINES   USING  GAS 

~Q2      wC*>  (T°  ~  r*)  ~  wC»  (T*  ~  r« 


E 


Qi  wCP  (Tc  - 

Cv   (Td-Tg\  i      T*  -  T 

~-*        ~ 


PROBLEMS 

1.  A  Carnot  engine  containing  10  Ibs.  of  air  has  at  the  beginning  of  the 
expansion  stroke  a  volume  of  10  cu.  ft.  and  a  pressure  of  200  Ibs.  per  sq.  in. 
absolute.    The  exhaust  temperature  is  o°  F.     If  10  B.t.u.  of  heat  are  added 
to  the  cycle,  find 

(a)  Efficiency  of  the  cycle. 

(b)  Work  of  the  cycle. 

2.  A  Carnot  cycle  has  at  the  beginning  of  the  expansion  stroke  a  pressure 
of  75  Ibs.  per  sq.  in.  absolute,  a  volume  of  2  cu.  ft.  and  a  temperature  of 
200°  F.    The  volume  at  the  end  of  the  isothermal  expansion  is  4  cu.  ft. 
The  exhaust  temperature  is  30°  F.     Find 

(a)  Heat  added  to  cycle. 

(b)  Efficiency  of  cycle. 

(c)  Work  of  cycle. 

3.  A  cycle  made  up  of  two  isothermal  and  two  adiabatic  curves  has  a 
pressure  of  100  Ibs.  per  sq.  in.  absolute  and  a  volume  of  i  cu.  ft.  at  the  begin- 
ning of  the  isothermal  expansion.     At  the  end  of  the  adiabatic  expansion 
the  pressure  is  10  Ibs.  per  sq.  in.  absolute  and  the  volume  is  8  cu.  ft.     Find 

(a)  Efficiency  of  cycle. 

(b)  Heat  added  to  cycle. 

(c)  Net  work  of  cycle. 

4.  In  a  Carnot  cycle  the  heat  is  added  at  a  temperature  of  400°  F.  and 
rejected  at  70°  F.    The  working  substance  is  i  Ib.  of  air  which  has  a  volume 
of  2  cu.  ft.  at  the  beginning  and  a  volume  of  4  cu.  ft.  at  the  end  of  the 
isothermal  expansion.     Find 

(a)  Volume  at  end  of  isothermal  compression. 

(b)  Heat  added  to  the  cycle. 

(c)  Heat  rejected  from  cycle. 

(d)  Net  work  of  the  cycle. 

5.  Air  at  a  pressure  of  100  Ibs.  per  sq.  in.  absolute,  having  a  volume  of 
i  cu.  ft.  and  a  temperature  of  200°  F.,  passes  through  the  following  opera- 
tions: 

i  st.   Heat  is  supplied  to  the  gas  while  expansion  takes  place  under 
constant  pressure  until  the  volume  equals  2  cu.  ft. 


PROBLEMS  6l 

2d.    It  then  expands  adiabatically  to  15  Ibs.  per  sq.  in.  absolute 
pressure. 

3d.    Heat  is  then  rejected  while  compression  takes  place  at  con- 
stant pressure. 

4th.  The  gas  is  then  compressed  adiabatically  to  its  original  volume 

of  i  cu.  ft. 
Find  (a)  Pounds  of  air  used. 

(b)  Temperature  at  end  of  constant  pressure  expansion. 

(c)  Heat  added  to  the  cycle. 

(d)  Net  work  of  cycle. 

(e)  Efficiency  of  cycle. 

6.  A  Stirling  hot-air  engine  having  a  piston  16  inches  in  diameter  and 
a  stroke  of  4  ft.  makes  28  r.p.m.     The  upper  temperature  is  650°  F.,  while 
the  lower  temperature  is  150°  F.     Assuming  that  the  volume  of  the  working 
cylinder  is  one-half  that  of  the  displacer  cylinder,  calculate  the  pressures  and 
volumes  at  each  point  of  the  cycle  (Fig.  8) ;  also  calculate  the  mean  effective 
pressure,  the  horse  power  developed,  and  the  cycle  efficiency. 

7.  Plot  the  Stirling  engine  cycle  from  the  results  of  problem  6. 

8.  Compare  the  cycle  efficiencies  of  internal-combustion  engines  working 
on  the  Otto  cycle  using  the  fuels  and  compression  pressures  as  indicated 
below: 

Gasoline,  75;  producer  gas,  150;  and  blast  furnace  gas,  200  Ibs.  per  sq.  in. 
gage  pressure. 

9.  Calculate  the  theoretical  pressures,  volumes  and  temperatures  at  each 
point  of  an  Otto  cycle,  as  well  as  the  mean  effective  pressure,  horse  power, 
and  cycle  efficiency  for  the  following  conditions  of  Fig.  n.    Assume  that 
the  pressure  after  compression  is  180  Ibs.  per  sq.  in.  gage,  that  80  B.t.u. 
are  added  during  combustion  and  that  n  in  PVn  =  1.4  : 

Pa  =  14-7  Ibs.  per  sq.  in.,  Va  =  13.5  cu.  ft., 
Ta  =  70  +  459-5  =  529-50  F.  absolute. 

10.  Prove  that  cycle  efficiency  of  the  Diesel  engine  depends  not  only 
upon  the  compression  pressure  before  ignition  but  also  upon  the  point  of 
cut-off  C  in  Fig.  13. 

11.  In  what  respects  do  the  actual  cycles  of  internal-combustion  engines 
differ  from  the  theoretical  Otto  and  Diesel  cycles? 

12.  The  demonstrations  in  the  text  for  the  various  gas  cycles  were  based 
on  the  assumption  that  the  specific  heats  of  gases  are  constant.    Since  the 
specific  heats  of  gases  vary  at  high  temperatures,  show  the  effect  of  this 
variability  on  the  cyclic  analysis.     Refer  to  Lucke's  Engineering  Thermo- 
dynamics, Levin's  Modern  Gas  Engines  and  the  Gas  Producer,  and  Clerk's 
Gas,  Petrol  and  Oil  Engines,  Vol.  II. 


CHAPTER  V 
PROPERTIES   OF  VAPORS 

Saturated  and  Superheated  Vapors.  As  was  explained  in  the 
chapter  on  the  properties  of  perfect  gases,  a  vapor  can  be  liqui- 
fied by  pressure  or  temperature  changes  alone.  At  every  pres- 
sure there  is  a  fixed  point,  called  the  point  of  vaporization,  at 
which  a  liquid  can  be  changed  into  a  vapor  by  the  addition  of 
heat.  A  vapor  near  the  point  of  vaporization  is  called  a  saturated 
vapor  and  has  a  definite  vaporization  temperature  for  any  given 
pressure.  When  a  vapor  is  heated  so  that  its  temperature  is 
greater  than  the  vaporization  temperature  corresponding  to  a 
given  pressure,  it  is  said  to  be  a  superheated  vapor.  Superheated 
vapors  only  when  far  removed  from  the  vaporization  tempera- 
ture approach  nearly  the  laws  of  perfect  gases. 

Theory  of  Vaporization.  When  heat  is  added  to  a  liquid  its 
temperature  will  rise  with  a  slight  volume  increase  until  the  point 
of  vaporization  is  reached.  This  is  always  a  definite  point  for 
any  given  pressure  and  depends  on  the  character  of  the  liquid. 
Thus  the  point  of  vaporization  of  water  at  the  atmospheric  pres- 
sure of  14.7  pounds  per  square  inch  is  212°  F.,  while  that  at  a  pres- 
sure of  150  pounds  absolute  is  358°  F.  On  the  other  hand,  the 
vaporization  temperature  of  ammonia  vapor  at  a  pressure  of  1 50 
pounds  absolute  is  79°  F.  The  increased  temperature  of  vapori- 
zation with  the  pressure  increase  is  due  to  the  fact  that  the  mole- 
cules being  crowded,  the  velocity  per  molecule,  or  temperature, 
must  be  great  enough  to  overcome  not  only  molecular  attraction 
but  also  external  pressure  before  vaporization  can  take  place. 

When  the  point  of  vaporization  is  reached,  any  further  addition 
of  heat  will  not  cause  any  temperature  increase,  but  internal 
work  accompanied  by  vaporization  will  be  produced  as  well  as 

an  enormous  increase  in  volume.     The  heat  required  to  entirely 

62 


VAPOR  TABLES  63 

vaporize  a  unit  weight  of  a  substance  at  a  given  pressure  is  termed 
the  latent  heat  of  vaporization. 

The  volume  developed  when  one  unit  weight  of  the  liquid  has 
been  completely  evaporated  is  called  the  specific  volume  of  the 
dry  vapor,  this  volume  depending  on  the  pressure. 

When  vaporization  is  incomplete  the  vapor  is  termed  a  wet 
saturated  vapor,  which  means  that  the  vapor  is  in  contact  with 
the  liquid  from  which  it  is  formed.  The  percentage  dryness  in 
the  vapor  is  called  its  quality.  Thus  the  quality  of  steam  is  0.97 
when  one  pound  of  it  consists  of  97  per  cent  steam  and  3  per  cent 
water.  While  the  volume  of  a  vapor  increases  with  the  quality, 
the  temperature  is  constant  between  the  point  of  vaporization 
and  the  point  of  superheat  for  any  given  pressure.  Thus  the 
temperature  of  steam  vapor  corresponding  to  150  pounds  abso- 
lute is  358°  F.,  no  matter  whether  the  quality  is  o.io,  0.75  or  i.oo. 

Further  addition  of  heat  after  complete  vaporization  will  cause 
temperature  as  well  as  volume  changes,  the  substance  being  in 
the  superheated  vapor  condition. 

Vapor  Tables.  The  exact  quantities  of  heat  required  to  pro- 
duce the  above  effects  under  various  conditions,  as  well  as  the 


220 


240  260  280 

Temperature,  Degrees  Fahrenheit 


300 


FIG.  14.  —  Pressure-Temperature  Relations  for  Saturated  Steam. 

relations  existing  between  pressure,  volume,  and  temperature  of 
saturated  and  superheated  vapors  have  been  determined  exper- 


64 


PROPERTIES  OF  VAPORS 


imentally.  These  experimental  results  have  been  given  in  the 
form  of  empirical  equations  from  which  vapor  tables  have  been 
computed.  Tables  3  and  4,  showing  the  properties  of  dry  satu- 


M  =  Marks  and  Davis 
G  =  Goodenough 
P  =•  Peabody 


12        14        16         18 
Volume,  cu.  ft.  per  Ib. 

FIG.  15.  —  Pressure-Volume  Relations  for  Saturated  Steam. 


Pressure,  Ibs.  per  sq.  in.,  Absolute 

8  S  §  §  1 

^ 

M-^^S 

X 

M  =  Mark 
G  =Gooc 
P  =  Peab 

s  and  Dai 
enough 
ody 

'is 

Py 

^N 

— 

X 

^ 

$^^ 

^ 

^^ 

880  900  920  940  960 

Latent  Heat 

FIG.  16.  —  Pressure-Latent  Heat  Relations  for  Saturated  Steam. 


rated  steam  and  of  ammonia,  are  given  in  the  appendix.  This 
text  should  be  supplemented  by  the  more  complete  tables  of 
Marks  and  Davis,  Peabody  or  Goodenough. 


TEMPERATURE,   PRESSURE  AND   VOLUME  OF   STEAM       65 

In  Figs.  14, 15  and  16  are  plotted  the  several  variables,  as  given 
in  the  above-mentioned  tables,  for  various  pressures. 

In  most  vapor  tables  will  be  found,  corresponding  to  the  pres- 
sure of  the  vapor  in  pounds  per  square  inch  absolute  (p),  the 
vaporization  temperature  (t),  the  heat  of  the  liquid  (q,  h  or  i'), 
the  heat  of  vaporization  (r  or  L),  the  specific  volume  (s  or  v)f 

entropy  of  water  (0,  5  or  n),  entropy  of  vaporization  ( —  or  —  V 

density  pounds  per  cubic  foot  (-  or  -J,  internal  latent  heat 

(p,  i  or  /).  In  some  tables  will  also  be  found  a  column  for  the 
total  heat  of  vapor  (H  =  q  +  r),  external  latent  heat  (APu) 
and  entropy. 

Relation  between  Temperature,  Pressure  and  Volume  of 
Saturated  Steam.  The  important  relations  of  temperature,  pres- 
sure and  volume  were  first  determined  in  a  remarkable  series  of 
experiments  conducted  by  a  French  engineer  named  Regnault, 
and  it  has  been  on  the  basis  of  his  data,  first  published  in  1847, 
that  even  our  most  modem  steam  tables  were  computed.  Later 
experimenters  have  found,  however,  that  these  data  were  some- 
what in  error,  especially  for  values  near  the  dry  saturated 
condition.  These  errors  resulted  because  it  was  difficult  in  the 
original  apparatus  to  obtain  steam  entirely  free  from  moisture. 

The  pressure  of  saturated  steam  increases  very  rapidly  as  the 
temperature  increases  in  the  upper  limits  of  the  temperature  scale. 
It  is  very  interesting  to  examine  a  table  of  the  properties  of 
steam  to  observe  how  much  more  rapidly  the  pressure  must  be 
increased  in  the  higher  limits  for  a  given  range  of  temperature. 

It  should  be  observed  that  in  most  tables  the  pressure  is  almost 
invariably  given  in  terms  of  pounds  per  square  inch,  while  in 
nearly  all  our  thermodynamic  calculations  the  pressure  must 
be  used  in  pounds  per  square  foot. 

The  specific  volume  of  saturated  steam  (v  or  s)  is  equal  to 
the  sum  of  the  volumes  of  water  (a)  and  of  the  increase  in 
volume  during  vaporization  (u),  or 

V  =  c  +  #. 


66  PROPERTIES  OF  VAPORS 

Heat  in  the  Liquid  (Water)  (h  or  q).  The  essentials  of  the 
process  of  making  steam  have  been  described  in  a  general  way. 
The  relation  of  this  process  to  the  amount  of  heat  required  will 
now  be  explained.  If  a  pound  of  water  which  is  initially  at 
some  temperature  to  is  heated  at  a  constant  pressure  P  (pounds 
per  square  foot)  to  the  boiling  point  corresponding  to  this  pres- 
sure and  then  converted  into  steam,  heat  will  first  be  absorbed 
in  raising  the  temperature  of  the  water  from  to  to  t,  and  then 
in  producing  vaporization.  During  the  first  stage,  while  the 
temperature  is  rising,  the  amount  of  heat  taken  in  is  approxi- 
mately (t  —  t0)  heat  units,  that  is,  British  thermal  units  (B.t.u.), 
because  the  specific  heat  of  water  is  approximately  unity  and 
practically  constant.  This  number  of  B.t.u.  multiplied  by  778 
gives  the  equivalent  number  of  foot-pounds  of  work.  For  the 
purpose  of  stating  in  steam  tables  the  amount  of  heat  required  for 
this  heating  of  water,  the  initial  temperature  to  must  be  taken  at 
some  definite  value;  for  convenience  in  numerical  calculations 
and  also  because  of  long  usage,  the  temperature  32°  F.  is  invari- 
ably used  as  an  arbitrary  starting  point  for  calculating  the 
amount  of  heat  "  taken  in."  The  symbol  h  (or  sometimes  i'  or  q) 
is  used  to  designate  the  heat  required  to  raise  one  pound  of 
water  from  32°  F.  to  the  temperature  at  which  it  is  vaporized  into 
steam.  In  other  words,  "the  heat  of  the  liquid"  (h)  is  the 
amount  of  heat  in  B.t.u.  required  to  raise  one  pound  of  water 
from  32°  F.  to  the  boiling  point,  or: 


h  =    I    C  dt.  (joi) 

C  is  the  specific  heat  of  water  at  constant  pressure,  /<,  is  the 
freezing  point  of  water,  /  is  the  temperature  to  which  the  water 
is  raised. 

As  C  is  very  nearly  unity,  the  value  of  the  heat  absorbed  by 
water  in  being  raised  to  the  steaming  temperature  (h)  in  B.t.u. 
can  be  expressed,  approximately,  by  the  formula 

h  -  t  -  32  (in  B.t.u.).  (102) 

Values  of  h,  taking  into  consideration  the  variation  in  the 


EXTERNAL  WORK  OF  EVAPORATION          67 

specific  heat  of  water,  will  be  found  in  the  usual  steam  tables 
(Table  3). 

During  this  first  stage,  before  any  steaming  has  occurred,  prac- 
tically all  the  heat  applied  is  used  to  increase  the  stock  of  internal 
energy.  The  amount  of  external  work  done  by  the  expansion 
of.  water  as  a  liquid  is  practically  negligible. 

Latent  Heat  of  Evaporation  (L  or  r).  In  the  second  stage  of  the 
formation  of  steam  as  described,  the  water  at  the  temperature  t, 
corresponding  to  the  pressure,  is  changed  into  steam  at  that 
temperature.  Although  there  is  no  rise  in  temperature,  very 
much  heat  is  nevertheless  required  to  produce  this  evaporation 
or  vaporization.  The  heat  taken  in  during  this  stage  is  the 
latent  heat  of  steam.  In  other  words,  the  latent  heat  of  steam 
may  be  defined  as  the  amount  of  heat  which  is  taken  in  by  a 
pound  of  water  while  it  is  changed  into  steam  at  constant  pres- 
sure, the  water  having  been  previously  heated  up  to  the  tem- 
perature at  which  steam  forms.  The  symbol  L  (also  sometimes 
r)  is  used  to  designate  this  latent  heat  of  steam.  Its  value 
varies  with  the  particular  pressure  at  which  steaming  occurs 
being  somewhat  smaller  in  value  at  high  pressures  than  at  low. 

The  latent  heat  (L  or  r)  of  saturated  steam  can  be  approxi- 
mately calculated  by  the  formula: 

L   =  970.4   —  0.655  (/  —   212)   —  0.00045  (/  —   2T2)2.      (103) 

External  Work  of  Evaporation  (Apu).  A  part  of  the  heat  taken 
in  during  the  "  steaming  "  process  is  spent  in  doing  external 
work.  Only  a  small  part  of  the  heat  taken  in  is  represented 
by  the  external  work  done  in  making  the  steam  in  the  boiler, 
and  the  remainder  of  the  latent  heat  (L)  goes  to  increase  the 
internal  energy  of  the  steam.  The  amount  of  heat  that  goes 
into  the  performing  of  external  work  is  equal  to  P  (the  pressure 
in  pounds  per  square  foot)  times  the  change  of  volume  (u) 
occurring  when  the  water  is  changed  into  steam,  or  Pu. 

Example.  At  the  usual  temperatures  of  the  working  fluid  in 
steam  engines  the  volume  of  a  pound  of  water  a  is  about  -gV  of  a 
cubic  foot.  The  external  work  done  in  making  one  pound  of 


68  PROPERTIES  OF  VAPORS 

steam  having  finally  a  volume  of  V  (cubic  feet)  at  a  constant  pres- 
sure P  (pounds  per  square  foot)  may  be  written  in  foot-pounds: 

External  work  =  Pu  =  P  (V  -  -gV).  (104) 

This  last  equation  can  be  expressed  in  British  thermal  units 
(B.t.u.)  by  APu  where  A  =  -^fg-.  It  is  apparent  also  from  this 
equation  that  the  external  work  done  in  making  steam  is  less  at 
low  pressure  than  at  high,*  because  there  is  less  resistance  to  over- 
come, or,  in  other  words,  P  in  equation  (104)  is  less.  The  heat 
equivalent  of  the  external  work  is,  therefore,  a  smaller  propor- 
tion of  the  heat  added  at  low  temperature  than  at  high. 

Total  Heat  of  Steam  (H).  The  heat  added  during  the  process 
represented  by  the  first  and  second  stages  in  the  formation  of  a 
pound  of  steam,  as  already  described,  is  called  the  total  heat  of 
saturated  steam  or,  for  short,  total  heat  of  steam,  and  is  repre- 
sented by  the  symbol  H.  Using  the  symbols  already  defined, 
we  can  write,  per  pound  of  steam, 

H  =  h  +  L  (B.t.u.).  (105) 

In  other  words,  this  total  heat  of  steam  is  the  amount  of  heat 
required  to  raise  one  pound  of  water  from  32°  F.  to  the  tempera- 
ture of  vaporization  and  to  vaporize  it  into  dry  steam  at  that 
temperature  under  a  constant  pressure. 

Remembering  that  h  for  water  is  approximately  equal  to  the 
temperature  less  32  degrees  corresponding  to  the  pressure  at 
which  the  steam  is  formed  (t)  the  total  heat  of  steam  is  approxi- 
mately, 

H  =  (t  -  32)  +  L.  (106) 

To  illustrate  the  application  of  equation  (106)  calculate  the 
total  heat  of  steam  being  formed  in  a  boiler  at  an  absolute  pres- 
sure of  115  pounds  per  square  inch. 

*  Although  at  the  lower  pressure  the  volume  of  a  given  weight  of  steam  is 
greater  than  at  a  higher  pressure,  the  change  of  pressure  is  relatively  so  much  greater 
in  the  process  of  steam  formation  that  the  product  of  pressure  and  change  of 
volume,  P  (Vz  —  Vi),  which  represents  the  external  work  done,  is  less  for  low 
pressure  steam  than  for  high. 


INTERNAL  ENERGY  OF  EVAPORATION  AND   OF   STEAM      69 

From  the  steam  tables  (Table  3)  the  temperature  t  of  the 
steam  at  a  pressure  of  115  pounds  per  square  inch  absolute  is 
338°  F.,  the  latent  heat  of  vaporization  is  880  B.t.u.  per  pound 
and  the  total  heat  of  the  steam  is  1189  B.t.u.  per  pound.  To 
check  these  values  with  equation  (106),  by  substituting  values  of 
L  and  t, 

H  =  (338  -  32)  +880  =  1186  B.t.u.  per  pound. 

When  steam  is  condensed  under  constant  pressure,  the  process 
which  was  called  the  "  second  stage  "  is  reversed  and  the  amount 
of  heat  equal  to  the  latent  heat  of  evaporation  (L)  is  given  up 
during  the  change  that  occurs  in  the  transformation  from  steam 
to  water. 

Internal  Energy  of  Evaporation  and  of  Steam.  It  was  ex- 
plained in  a  preceding  paragraph  that  when  steam  is  forming 
not  all  of  the  heat  added  goes  into  the  internal  or  "  intrinsic  " 
energy  of  the  steam,  but  that  a  part  of  it  was  spent  in  performing 
external  work.  If,  then,  the  internal  energy  of  evaporation  is 
represented  by  the  symbol  IL, 

.  (107) 


This  equation  represents  the  increase  in  internal  energy  which 
takes  place  in  the  changing  of  a  pound  of  water  at  the  tem- 
perature t  into  steam  at  the  same  temperature.  In  all  the 
formulas  dealing  with  steam  the  state  of  water  at  32°  F.  has  been 
adopted  as  the  arbitrary  starting  point  from  which  the  taking  in 
of  heat  was  calculated.  This  same  arbitrary  starting  point  is 
used  also  in  expressing  the  amount  of  internal  energy  in  the  steam. 
This  is  the  excess  of  the  heat  taken  in  over  the  external  work  done 
in  the  process. 

The  total  internal  energy  (1H)  of  a  pound  of  saturated  steam  at 
a  pressure  P  in  pounds  per  square  foot  is  equal  to  the  total  heat 
(H)  less  the  heat  equivalent  of  the  external  work  done;  thus, 

1H  =  H  -P(V  -f*^  (108) 

778 


70  PROPERTIES  OF  VAPORS 

Such  reference  is  made  here  to  the  internal  energy  of  steam 
because  it  is  very  useful  in  calculating  the  heat  taken  in  and 
rejected  by  steam  during  any  stage  of  its  expansion  or  com- 
pression. 

Heat  taken  in  =  increase  of  internal  energy  +  external  work 
done. 

When  dealing  with  a  compression  instead  of  an  expansion  the 
last  term  above  (external  work)  will  be  a  negative  value  to  indi- 
cate that  work  is  done  upon  the  steam  instead  of  the  steam  doing 
work  by  expansion. 

The  following  problem  shows  the  calculation  of  internal  energy 
and  external  work: 

Example.  A  boiler  is  evaporating  water  into  dry  and  satu- 
rated steam  at  a  pressure  of  300  pounds  per  square  inch  abso- 
lute. The  feed  water  enters  the  boiler  at  a  temperature  of  145°  F. 

The  internal  energy  of  evaporation  per  .pound  of  steam  is 

,        T       P  (V  -  TJV)       -  300  X  144  (i.55i  ~  iftr) 

IL  =  Li =  'oil.  3 

778  778 

=  811.3  —  85.3  =  726.0  B.t.u., 

or  taken  directly  from  saturated  steam  tables  equals  726.8  B.t.u. 
The  total  internal  energy  supplied  above  32°  F.  per  pound  of 
steam  is 

In  =  H  -  P(y  ~  ^  =  1204.1  -  85.3  =  1118.8  B.t.u., 

778 

or  taken  directly  from  saturated  steam  tables  equals  1118.5  B.t.u. 
External  work  done  above  32°  F.  per  pound  of  steam  as  calcu- 
lated from  steam  tables  is 

H  —  IH  =  1204.1  —  1118.5  =  85-6  B.t.u. 

External  work  of  evaporation  per  pound  of  steam  as  calcu- 
lated from  the  steam  tables  is 

L  -  IL  =  811.3  -  726.8  =  84.5  B.t.u. 

The  external  work  done  in  raising  the  temperature  of  a  pound 
of  water  from  32°  F.  to  the  boiling  point  is 

85.6  -  84.5  =  i.i  B.t.u. 


WET   STEAM  71 

The  external  work  done  in  raising  the  temperature  of  a  pound 
of  water  from  32°  F.  to  145°  F.  is 

300  X  144  (0.0163  -  0.01602)  = 
778 

The  external  work  done  in  forming  the  steam  from  water  at 
145°  F.  is,  then, 

84.5  +  1.  10  —  o.oi  =  85.59  B.t.u. 
as  compared  with  85.3  B.t.u.  as  calculated  from 

P(V  -  irV) 
778 

It  is  to  be  noticed  that  the  external  work  done  during  the  addi- 
tion of  heat  when  the  liquid  is  raised  to  the  vaporization  tem- 
perature is  small  as  compared  with  other  values,  and  for  most 
engineering  work  it  is  customary  to  assume  that  no  external  work 
is  done  during  the  addition  of  heat  to  the  water.  With  this 
assumption 

P(V  ~  *V) 


775 


(109) 


Steam  Formed  at  Constant  Volume.  When  saturated  steam 
is  made  in  a  boiler  at  constant  volume,  as,  for  example,  when  the 
piping  connections  from  the  boiler  to  the  engines  are  closed,  then 
no  external  work  is  done,  and  all  the  heat  taken  in  is  converted 
into  and  appears  as  internal  energy  IH  of  the  steam.  This 
quantity  is  less  than  the  total  heat  H  of  steam,  representing  its 
formation  at  constant  pressure,  by  quantity  P  (V  —  ^V)  •*-  778, 
where  P  represents  the  absolute  pressure  at  which  the  steam  is 
formed  in  pounds  per  square  foot  and  V  is  the  volume  of  a 
pound  of  steam  at  this  pressure  in  cubic  feet. 

Wet  Steam.  In  all  problems  studied  thus  far,  dealing  with 
saturated  steam,  it  has  been  assumed  that  the  steaming  process 
was  complete  and  that  the  water  had  been  completely  converted 
into  steam.  In  actual  practice  it  is  not  at  all  unusual  to  have 
steam  leaving  boilers  which  is  not  perfectly  and  completely  vapor- 


72  PROPERTIES  OF  VAPORS 

ized;  in  other  words,  the  boilers  are  supplying  to  the  engines  a 
mixture  of  steam  and  water.  This  mixture  we  call  wet  steam. 
It  is  steam  which  carries  actually  in  suspension  minute  particles 
of  water,  which  remain  thus  in  suspension  almost  indefinitely. 
The  temperature  of  wet  steam  is  always  the  same  as  that  of  dry 
saturated  steam  as  given  in  the  steam  tables,  so  long  as  any 
steam  remains  uncondensed. 

The  ratio  of  the  weight  of  moisture  or  water  in  a  pound  of  wet 
steam  to  a  pound  of  completely  saturated  steam  is  called  the 
degree  of  wetness;  and  when  this  ratio  is  expressed  as  a  per 
cent,  it  is  called  the  percentage  of  moisture  or  "  per  cent  wet." 
If  in  a  pound  of  wet  steam  there  is  0.04  pound  of  water  in  suspen- 
sion or  entrained,  the  steam  is  four  per  cent  wet. 

Another  term,  called  the  quality  of  steam,  which  is  usually 
expressed  by  the  symbol  x,  is  also  frequently  used  to  represent 
the  condition  of  wet  steam.  Quality  of  steam  may  be  defined 
as  the  proportion  of  the  amount  of  dry  or  completely  evaporated 
steam  in  a  pound  of  wet  steam.  To  illustrate  with  the  example 
above,  if  there  is  0.04  pound  of  water  in  a  pound  of  wet  steam; 
the  quality  in  this  case  would  be  i  —  0.04  or  0.96.  In  this  case 
the  quality  of  this  steam  is  ninety-six  one-hundredths. 

With  this  understanding  of  the  nature  of  wet  steam  it  is  obvi- 
ous that  latent  heat  of  a  pound  of  wet  steam  is  xL.  Similarly, 
the  total  heat  of  a  pound  of  wet  steam  is  h  +  xL,  and  the 
volume  of  a  pound  of  wet  steam  is  x  (V  —  A)  +  A  =  xV  +  A 
(i  —  x),  or  approximately  it  is  equal  to  xV,  because  the  term  A 
(i  —  x)  is  negligibly  small  except  in  cases  where  the  steam  is  so 
wet  as  to  consist  mostly  of  water.  Similarly,  the  internal  energy 
in  a  pound  of  wet  steam  is 


Superheated  Steam.  When  the  temperature  of  steam  is  higher 
than  that  corresponding  to  saturation  as  taken  from  the  steam 
tables,  and  is,  therefore,  higher  than  the  standard  temperature 
corresponding  to  the  pressure,  the  steam  is  said  to  be  super- 


THE  TOTAL  HEAT  OF  SUPERHEATED  STEAM      73 

heated.  In  this  condition  steam  begins  to  depart  and  differ 
from  its  properties  in  the  saturated  condition,  and  when  super- 
heated to  a  very  high  degree  it  begins  to  behave  somewhat 
like  a  perfect  gas. 

There  are  tables  of  the  properties  of  superheated  steam  just 
as  there  are  tables  of  saturated  steam.  (See  Marks  and  Davis', 
Goodenough's  or  Peabody's  Steam  Tables.)  When  dealing  with 
saturated  steam  there  is  always  only  one  possible  temperature  and 
only  one  specific  volume  to  be  considered.  With  superheated 
steam,  on  the  other  hand,  for  a  given  pressure  there  may  be  any 
temperature  above  that  of  saturated  steam,  and  corresponding  to 
each  temperature  there  will  be,  of  course,  definite  values  for 
specific  volume  and  total  heat.  Like  a  perfect  gas  the  specific 
volume  or  the  cubic  feet  per  pound  increases  with  the  increase  in 
temperature. 

The  total  heat  of  superheated  steam  is  obviously  greater  than 
the  total  heat  of  saturated  steam.*  Thus,  since  the  total  heat 
of  dry  saturated  steam  is,  as  before,  h  +  Z,,  the  total  heat  of 
superheated  steam  with  D  degrees  of  superheat  is 

Hs  =  h  +  L  +  CP  x  Z>;  (in) 

or  if  the  temperature  of  the  superheated  steam  is  feuP.  and  A»t.  is 
the  temperature  of  saturated  steam  corresponding  to  the  pressure, 

Us   =  h  +  L  +  Cp  (/sup.   -  /sat.).  (112) 

Hs  (equations  in  and  112)  is  the  total  heat  of  superheated 
steam  or  is  the  amount  of  heat  required  to  produce  a  pound  of 
steam  with  the  required  degrees  of  superheat  from  water  at  32°  F. 

To  obtain  the  amount  of  internal  energy  of  a  pound  of  steam 
(superheated)  corresponding  to  this  total  heat  as  stated  above, 
the  external  work  expended  in  the  steaming  process  must  be 
subtracted;  that  is, 

*  In  practice  steam  is  called  dry  saturated  when  it  is  exactly  saturated  and  has 
no  moisture.  It  is  the  condition  known  simply  as  saturated  steam  as  regards  the 
properties  given  in  the  ordinary  steam  tables.  The  dry  saturated  condition  is  the 
boundary  between  wet  steam  and  superheated  steam. 


74  PROPERTIES   OF   VAPORS 

IH  =  h  +  L  -  P  (V  ~-  A)  +  C 


. 


. 

77° 

=  H,-P(V™-*>.  (113) 

The  term  g»  can  be  neglected  in  the  equations  above  for  prac- 
tically all  engineering  calculations  as  the  maximum  error  frora 
this  is  not  likely  to  be  more  than  one  in  one  thousand  or  TO  per 
cent.*  The  accuracy  of  our  steam  tables  for  values  of  latent  and 
total  heat  is  not  established  to  any  greater  degree.  Making 
this  approximation,  the  equation  above  becomes 

IZ,     I      T      I     r>     ft  -t       ^  I  •*      *     r  supA          TJ  P  X    V  sap. 

H  -  n  +  L  +  Cp  (Up.  -  /sat.)  -  (  —  —  -  —  1  =  H,  - 

If  examination  is  made  of  the  properties  of  superheated  steam 
as  given  in  Marks  and  Davis'  or  in  Goodenough's  Steam  Tables 
for  superheated  steam  at  165  pounds  per  square  inch  absolute 
pressure  and  150°  F.  superheat,  the  following  results  are  secured: 

Marks  &  Davis       Goodenough 

Temperature  (/)  ...................        516.0  516.1 

Volume  (V,v)  .....................  343  34* 

Total  Heat  (H,  or  Af  or  i)  .....  ......      1277.6  1280.8 

The  values  of  h  and  L  for  saturated  steam  at  this  pressure  are 
respectively  338.2  and  856.8.  From  the  curves  given,  Fig.  17,^ 

*  The  error  in  the  value  of  total  internal  energy  due  to  neglecting  the  term  -jfo 
in  the  exercise  worked  out  on  page  70  is  0.85  B.t.u.  per  pound  (one  per  cent  of  the 
external  work),  or  an  error  of  .076  per  cent  in  the  final  result. 

f  In  their  tables  Marks  and  Davis  represent  the  total  heat  of  superheated 
steam  by  h.  In  this  book  the  symbol  Hs  is  used  for  greater  clearness  and  to  be 
consistent  with  the  symbols  used  in  the  preceding  formulas.  They  use  also  v  for 
specific  volume  in  place  of  V  as  above. 

J  The  curves  for  values  of  Cp,  as  given  in  Fig.  17,  are  for  average  and  not  for 
instantaneous  values  such  as  are  given  in  Fig.  18,  of  Marks  and  Davis'  Tables  and 
Diagrams.  Great  caution  must  be  observed  in  the  use  of  curves  of  this  kind. 
Those  giving  instantaneous  values  can  only  be  used  for  the  value  of  Cp  in  the  form- 
ulas given  for  the  total  heat  of  superheated  steam  after  the  average  value  has  been 
found  by  integrating  the  curve  representing  these  values  for  a  given  pressure. 


DRYING  OF   STEAM   BY  THROTTLING  OR  WIRE-DRAWING      75 


the  specific  heat  of  superheated  steam  at  constant  pressure  (Cp) 
is  0.552.     From  these  data, 
Hs  =  h  +  L  +  CP  X  150  =  338.2  +  856.8  +  0.552  X  150  =  1277.7. 

This  value  agrees  very  nearly  with  that  given  in  the  tables  as 
previously  indicated. 


.42 
200    250    300   350    400 

Temperature  °F 

FIG.  17.  —  Mean  Values  of  Cp  Calculated  by  Integration  from  Knoblauch 
and  Jakob's  Data. 

The  specific  volume  (F)  is  calculated  from  the  empirical  form- 
ula derived  from  experimental  results  and  is  expressed  as  follows: 


T-p  (1+0.0014 


(114) 


where  p  is  in  pounds  per  square  inch,  F  is  in  cubic  feet  per  pound 
and  T  =  t  +  460  is  the  absolute  temperature  on  the  Fahrenheit 
scale.* 

Drying  of  Steam  by  Throttling  or  Wire-drawing.     When  steam 
expands  by  passing  through  a  very  small  opening,  as,  for  example, 

^1 

*  The  value  of  7  or  -~   of  superheated  steam  used  in  ordinary  engineering 
calculations  is  1.3. 


76 


PROPERTIES  OF  VAPORS 


through  a  valve  only  partly  open  in  a  steam  line,  the  pressure 
is  considerably  reduced.  This  effect  is  called  throttling  or  wire- 
drawing. The  result  of  expansion  of  this  kind  when  the  pressure 


0.90 


-200  250  o  300 

temperature    G. 

FIG.  18.  —  Values  of  the  "True"  Specific  Heat  of  Superheated  Steam. 

is  reduced  and  no  work  is  done  is  that  if  the  steam  is  initially 
wet  it  will  be  drier  and  if  it  is  initially  dry  or  superheated  the 
degree  of  superheat  will  be  increased.  The  reason  for  this  is 
that  the  total  heat  required  to  form  a  pound  of  dry  saturated 
steam  (H)  is  considerably  less  at  low  pressure  than  at  high,  but 
obviously  the  total  quantity  of  heat  in  a  pound  of  steam  must  be 


THROTTLING  OR  SUPERHEATING   CALORIMETERS  77 

the  same  after  wire-drawing  as  it  was  before,  neglecting  radiation. 
Now  if  steam  is  initially  wet  and  the  quality  is  represented  by 
Xi,  then  the  total  heat  in  the  steam  is  represented  by  hi  plus  xiLi, 
in  which  hi  and  LI  represent  the  heat  of  the  liquid  and  the  latent 
heat  of  the  steam  at  the  initial  pressure.  If,  also,  the  quality, 
heat  of  liquid  and  the  latent  heat  of  the  steam  after  wire-drawing 
are  represented  respectively  by  #2,  fe  and  Z^,  then  hi  +  x\L\ 
=  fa  +  «aL2,  or 

ft  = XlLl  +£  ~  fe-  (us) 

This  drying  action  of  steam  in  passing  through  a  small  open- 
ing or  an  orifice  is  very  well  illustrated  by  steam  discharging 
from  a  small  leak  in  a  high  pressure  boiler  into  the  atmosphere. 
It  will  be  observed  that  no  moisture  is  visible  in  the  steam  a 
few  inches  from  the  leak  but  farther -off  it  becomes  condensed 
by  loss  of  heat,  becomes  clouded  and  plainly  visible.  An  im- 
portant application  of  the  throttling  principle  is  also  to  be  found 
in  the  throttling  calorimeter  (page  80). 

Determination  of  the  Moisture  in  Steam.  Unless  the  steam 
used  in  the  power  plant  is  superheated  it  is  said  to  be  either  dry 
or  wet,  depending  on  whether  or  not  it  contains  water  in  sus- 
pension. The  general  types  of  steam  calorimeters  used  to  de- 
termine the  amount  of  moisture  in  the  steam  may  be  classified 
under  three  heads: 

1 .  Throttling  or  superheating  calorimeters. 

2.  Separating  calorimeters. 

3.  Condensing  calorimeters. 

Throttling  or  Superheating  Calorimeters.  The  type  of  steam 
calorimeter  used  most  in  engineering  practice  operates  by  passing 
a  sample  of  the  steam  through  a  very  small  orifice,  in  which  it 
is  superheated  by  throttling.  A  very  satisfactory  calorimeter  of 
this  kind  can  be  made  of  pipe  fittings  as  illustrated  in  Fig.  19. 
It  consists  of  an  orifice  O  discharging  into  a  chamber  C,  into 
which  a  thermometer  T  is  inserted,  and  a  mercury  manometer 


78  PROPERTIES  OF  VAPORS 

is  usually  attached  to  the  cock  V3  for  observing  the  pressure  in 
the  calorimeter. 

It  is  most  important  that  all  parts  of  calorimeters  of  this 
type,  as  well  as  the  connections  leading  to  the  main  steam  pipe, 
should  be  very  thoroughly  lagged  by  a  covering  of  good  insu- 
lating material.  One  of  the  best  materials  for  this  use  is  hair 
felt,  and  it  is  particularly  well  suited  for  covering  the  more  or 
less  temporary  pipe  fittings,  valves  and  nipples  through  which 
steam  is  brought  to  the  calorimeter.  Throttling  calorimeters 
have  been  found  useless  because  the  small  pipes  leading  to  the 
calorimeters  were  not  properly  lagged,  so  that  there  was  too 
much  radiation,  producing,  of  course,  condensation,  and  the 
calorimeter  did  not  get  a  true  sample.  It  is  obvious  that  if  the 
entering  steam  contains  too  much  moisture  the  drying  action 
due  to  the  throttling  in  the  orifice  may  not  be  sufficient  to  super- 
heat. It  may  be  stated  in  general  that  unless  there  are  about 
5°  to  10°  F.  of  superheat  in  the  calorimeter,  or,  in  other  words, 
imless  the  temperature  on  the  low  pressure  side  of  the  orifice  is 
at  least  about  5°  to  10°  F.  higher  than  that  corresponding  to  the 
pressure  in  the  calorimeter,  there  may  be  some  doubt  as  to  the 
accuracy  of  results.*  The  working  limits  of  throttling  calo- 
rimeters vary  with  the  initial  pressure  of  the  steam.  For  35 
pounds  per  square  inch  absolute  pressure  the  calorimeter  ceases 
to  superheat  when  the  percentage  of  moisture  exceeds  about 
2  per  cent;  for  150  pounds  absolute  pressure  when  the  moisture 
exceeds  about  5  per  cent;  and  for  250  pounds  absolute  pressure 
when  it  is  in  excess  of  about  7  per  cent.  For  any  given  pressure 
the  exact  limit  varies  slightly,  however,  with  the  pressure  in  the 
calorimeter. 

*  The  same  general  statement  may  be  made  as  regards  determinations  of 
superheat  in  engine  and  turbine  tests.  Experience  has  shown  that  tests  made 
with  from  o  to  10  degrees  Fahrenheit  superheat  are  not  reliable,  and  that  the 
steam  consumption  in  many  cases  is  not  consistent  when  compared  with  results 
obtained  with  wet  or  more  highly  superheated  steam.  The  errors  mentioned 
when  they  occur  are  probably  due  to  the  fact  that  in  steam,  indicating  less  than 
10  degrees  Fahrenheit  superheat,  water  in  the  liquid  state  may  be  taken  up  in 
"slugs"  and  carried  along  without  being  entirely  evaporated. 


THROTTLING  OR   SUPERHEATING  CALORIMETERS  79 

In  connection  with  a  report  on  the  standardizing  of  engine 
tests,  the  American  Society  of  Mechanical  Engineers  *  published 
instructions  regarding  the  method  to  be  used  for  obtaining  a 
fair  sample  of  the  steam  from  the  main  pipes.  It  is  recom- 
mended in  this  report  that  the  calorimeter  shall  be  con- 
nected with  as  short  intermediate  piping  as  possible  with  a 
so-called  calorimeter  nipple  made  of  J-inch  pipe  and  long  enough 
to  extend  into  the  steam  pipe  to  within  |  inch  of  the  opposite 
wall.  The  end  of  this  nipple  is  to  be  plugged  so  that  the  steam 
must  enter  through  not  less  than  twenty  |-inch  holes  drilled 
around  and  along  its  length.  None  of  these  holes  shall  be  less 
than  \  inch  from  the  inner  side  of  the  steam  pipe.  The  sample 
of  steam  should  always  be  taken  from  a  vertical  pipe  as  near  as 
possible  to  the  engine,  turbine  or  boiler  being  tested.  A  good 
example  of  a  calorimeter  nipple  is  illustrated  in  Fig.  20. 

The  discharge  valve  V2  should  not  be  closed  or  adjusted 
without  first  closing  the  gage  cock  V3.  Unless  this  precaution  is 
taken,  the  pressure  may  be  suddenly  increased  in  chamber  C,  so 
that  if  a  manometer  is  used  the  mercury  will  be  blown  out  of  it; 
and  if,  on  the  other  hand,  a  low-pressure  steam  gage  is  used  it 
may  be  ruined  by  exposing  it  to  a  pressure  much  beyond  its  scale. 

Usually  it  is  a  safe  rule  to  begin  to  take  observations  of  tem- 
perature in  calorimeters  after  the  thermometer  has  indicated  a 
maximum  value  and  has  again  slightly  receded  from  it.  The 
quality  or  relative  dryness  of  wet  steam  is  easily  calculated  by 
the  following  method  and  symbols : 

pi  =  steam  pressure  in  main,  pounds  per  square  inch 

absolute, 
pz  =  steam  pressure  in  calorimeter,  pounds  per  square 

inch  absolute, 

tc  =  temperature  in  calorimeter,  degrees  Fahrenheit, 
Li  and  hi  =  heat  of  vaporization  and   heat  of  liquid  corre- 
sponding to  pressure  pi,  B.t.u., 
Hz  and  /2  =  total    heat    (B.t.u.)    and  temperature    (degrees 

Fahrenheit)  corresponding  to  pressure  />2, 
*  Proceedings  American  Society  of  Mechanical  Engineers,  vol.  XXI. 


8o 


PROPERTIES  OF  VAPORS 


Cp  =  specific  heat  of  superheated  steam.    Assume  0.47 

for  low  pressures  existing  in  calorimeters, 
Xi  =  initial  quality  of  steam. 


nection  for 
Manometer  '" 


FIG.  19.  —  Simple  Throttling  Steam  Calorimeter. 

Total  heat  in  a  pound  of  wet  steam  flowing  into  orifice  is 


and  after  expansion,  assuming  all  the  moisture  is  evaporated, 
the  total  heat  of  the  same  weight  of  steam  is 

H2  +  CP(tc  -fe). 

Then  assuming  no  heat  losses  and  putting  for  CP  its  value  0.47  : 
fe  =  #2  +  0.47  (tc  -  fe)  (n6) 

-  #2  +  0-47  (fe  ~  fe)  ~  h 
Xi  - 


BARRUS  THROTTLING   CALORIMETER  8l 

The  following  example  shows  the  calculations  for  finding  the 
quality  of  steam  from  the  observations  taken  with  a  throttling 
calorimeter: 

Example.  Steam  at  a  pressure  of  100  pounds  per  square  inch 
absolute  passes  through  a  throttling  calorimeter.  In  the  calo- 
rimeter the  temperature  of  the  steam  becomes  243°  F.  and  the 
pressure  15  pounds  per  square  inch  absolute.  Find  the  quality. 

Solution.  By  taking  values  directly  from  tables  of  properties 
of  superheated  steam,*  the  total  heat  of  the  steam  in  the  calorim- 
eter at  15  pounds  per  square  inch  absolute  pressure  and  243°  F. 
is  1164.8  B.t.u.  per  pound  or  can  be  calculated  as  follows: 

E  +  Cp  (tc  -  fe)  =  1150.7  +  0-47  (243  -  2I3) 
=  1164.8  B.t.u.  per  pound. 

(Note.  tc  =  temperature  in  calorimeter  and  k  =  temperature 
corresponding  to  calorimeter  pressure.) 

The  total  heat  of  the  steam  before  entering  the  calorimeter 
is  h  H-  xL.  At  100  pounds  per  square  inch  absolute  pressure, 
this  is  298.3  -h  888.0  x  in  B.t.u.  per  pound.  Since  the  heat  in 
the  steam  per  pound  in  the  calorimeter  is  obviously  the  same  as 
before  it  entered  the  instrument,  we  can  equate  as  follows : 

298.3  +  888.0  #  =  1164.8 

x  =  0.976 
or  the  steam  is  2.4  per  cent- wet. 

Barrus  Throttling  Calorimeter.  This  is  an  important  varia- 
tion from  the  type  of  throttling  calorimeter  shown  in  Fig.  19  and 
has  been  quite  widely  introduced  by  Mr.  George  H.  Barrus.  In 
this  apparatus  the  temperature  of  the  steam  admitted  to  the 
calorimeter  is  observed  instead  of  the  pressure  and  a  very  free 
exhaust  is  provided  so  that  the  pressure  in  the  calorimeter  is 
atmospheric.  This  arrangement  simplifies  very  much  the  obser- 
vations to  be  taken,  as  the  quality  of  the  steam  x\  can  be  calcu- 
lated by  equation  (117)  by  observing  only  the  two  temperatures 

*  Marks  and  Davis'  Steam  Tables  and  Diagrams  (ist  ed.),  page  24  or 
Goodenough's  Properties  of  Steam  and  Ammonia  (26.  ed.),  page  47. 


82 


PROPERTIES   OF  VAPORS 


ti  and  tC)  taken  respectively  on  the  high  and  low  pressure  sides  of 
the  orifice  in  the  calorimeter.  This  calorimeter  is  illustrated  in 
Fig.  20.  The  two  thermometers  required  are  shown  in  the 
figure.  Arrows  indicate  the  path  of  the  steam.* 

The  orifice  in  such  calorimeters  is  usually  made  about  ^\  inch 
in  diameter,  and  for  this  size  of  orifice  the  weight  of  steam  | 

discharged  per  hour  at  175  pounds 
per  square  inch  absolute  pressure 
is  about  60  pounds.  It  is  impor- 
tant that  the  orifice  should  always 
be  kept  clean,  because  if  it  be- 
comes obstructed  there  will  be  a 
reduced  quantity  of  steam  passing 
through  the  instrument,  making 
the  error  due  to  radiation  rela- 
tively more  important. 

In  order  to  free  the  orifice  from 


1. 

/ 
/ 

I 

' 

/ 
/ 

FIG.  20.  —  Barms'  Throttling  Steam 
Calorimeter. 


dirt  or  other  obstructions  the  connecting  pipe  to  be  used  for 
attaching  the  calorimeter  to  the  main  steam  pipe  should  be 
blown  out  thoroughly  with  steam  before  the  calorimeter  is  put 
in  place.  The  connecting  pipe  and  valve  should  be  covered 
with  hair  felting  not  less  than  f  inch  thick.  It  is  desirable  also 
that  there  should  be  no  leak  at  any  point  about  the  apparatus, 
as  in  the  stuffing-box  of  the  supply  valve,  the  pipe  joints  or  the 
union. 

With  the  help  of  a  diagram^  giving  the  quality  of  steam  directly 
the  Barrus  calorimeter  is  particularly  well  suited  for  use  in 
power  plants,  where  the  quality  of  the  steam  is  entered  regu- 
larly on  the  log  sheets.  The  percentage  of  moisture  is  obtained 
immediately  from  two  observations  without  any  calculations. 

Separating  Calorimeters.  It  was  explained  that  throttling 
calorimeters  cannot  be  used  for  the  determination  of  the  quality 
of  steam  when  for  comparatively  low  pressures  the  moisture 

*  Transactions  of  American  Society  of  Mechanical  Engineers,  vol.  XI,  page  790. 
t  In  boiler  tests  corrections  should  be  made  for  the  steam  discharged  from  the 
steam  calorimeters. 

|  See  page  113,  and  also  Mover's  Power  Plant  Testing,  (2d  ed.),  pages  58-60. 


SEPARATING   CALORIMETERS 


is  in  excess  of  2  per  cent,  and  when  for  boiler  pressures  commonly 
used  it  exceeds  5  per  cent.  For  higher  percentages  of  moisture 
than  these  low  limits  separating  calorimeters  are  most  generally 
used.  In  these  instruments  the  water  is  removed  from  the 
sample  of  steam  by  mechanical  separation  just  as  it  is  done 
in  the  ordinary  steam  separator  installed  in  the  steam  mains 
of  a  power  plant.  There  is  pro- 
vided, of  course,  a  device  for  de- 
tenriining,  while  the  calorimeter  is 
in  operation,  usually  by  means  of 
a  calibrated  gage  glass,  the  amount 
of  moisture  collected.  This  me- 
chanical separation  depends  for  its 
action  on  changing  very  abruptly 
the  direction  of  flow  and  reducing 
the  velocity  of  the  wet  steam. 
Then,  since  the  moisture  (water) 
is  nearly  300  times  as  heavy  as 
steam  at  the  usual  pressures  de- 
livered to  the  engine,  the  moisture 
will  be  deposited  because  of  its 
greater  inertia. 

Fig.  21  illustrates  a  form  of  sep- 
arating calorimeter  having  a  steam    FlG>  2I.  _  separating  Calorimeter, 
jacketing  space  which  receives  live 

steam  at  the  same  temperature  as  the  sample.  Steam  is  supplied 
through  a  pipe  A,  discharging  into  a  cup  B.  Here  the  direction 
of  the  flow  is  changed  through  nearly  180  degrees,  causing  the 
moisture  to  be  thrown  outward  through  the  meshes  in  the  cup 
into  the  vessel  V.  The  dry  steam  passes  upward  through  the 
spaces  between  the  webs  W,  into  the  top  of  the  outside  jacketing 
chamber  J,  and  is  finally  discharged  from  the  bottom  of  this 
steam  jacket  through  the  nozzle  N.  This  nozzle  is  consider- 
ably smaller  than  any  other  section  through  which  the  steam 
flows,  so  that  there  is  no  appreciable  difference  between  the  pres- 
sures in  the  calorimeter  proper  and  the  jacket.  The  scale 


84  PROPERTIES  OF  VAPORS 

opposite  the  gage  glass  G  is  graduated  to  show,  in  hundredths 
of  a  pound,  at  the  temperature  corresponding  to  steam  at 
ordinary  working  pressures,  the  variation  of  the  level  of  the 
water  accumulating.  A  steam  pressure  gage  P  indicates  the 
pressure  in  the  jacket  J,  and  since  the  flow  of  steam  through 
the  nozzle  N  is  roughly  proportional  to  the  pressure,  another 
scale  in  addition  to  the  one  reading  pressures  is  provided 
at  the  outer  edge  of  the  dial.  A  petcock  C  is  used  for  draining 
the  water  from  the  instrument,  and  by  weighing  the  water  col- 
lected corresponding  to  a  given  difference  in  the  level  in  the  gage 
G'  the  graduated  scale  can  be  readily  calibrated.  Too  much 
reliance  should  not  be  placed  on  the  readings  for  the  flow  of  steam 
as  indicated  by  the  gage  P  unless  it  is  frequently  calibrated. 
Usually  it  is  very  little  trouble  to  connect  a  tube  to  the  nozzle  N 
and  condense  the  steam  discharged  in  a  large  pail  nearly  filled 
with  water.  When  a  test  for  quality  is  to  be  made  by  this  method 
the  pail  nearly  filled  with  cold  water  is  carefully  weighed,  and  then 
at  the  moment  when  the  level  of  the  water  in  the  water  gage  G 
has  been  observed  the  tube  attached  to  the  nozzle  N  is  immedi- 
ately placed  under  the  surface  of  the  water  in  the  pail.  The 
test  should  be  stopped  before  the  water  gets  so  hot  that  some 
weight  is  lost  by  "steaming."  The  gage  P  is  generally  cali- 
brated to  read  pounds  of  steam  flowing  in  ten  minutes.  For  the 
best  accuracy  it  is  desirable  to  use  a  pail  with  a  tightly  fitting 
cover  into  which  a  hole  just  the  size  of  the  tube  has  been  cut. 

If  W  is  the  weight  of  dry  steam  flowing  through  the  orifice  N 
and  w  is  the  weight  of  moisture  separated,  the  quality  of  the 

steam  is 

W 

1  x=^-  I  || (iis) 

Condensing  or  Barrel  Calorimeter.  For  steam  having  a  large 
percentage  of  moisture  (over  5  per  cent)  the  condensing  or 
barrel  calorimeter  will  give  fairly  good  results  if  properly  used. 
In  its  simplest  form  it  consists  of  a  barrel  placed  on  a  platform 
scale  and  containing  a  known  weight  of  cold  water.  The  steam 
is  introduced  by  a  pipe  reaching  nearly  to  the  bottom  of  the 


EQUIVALENT  EVAPORATION  AND  FACTOR  OF  EVAPORATION   85 

barrel.  The  condensation  of  the  steam  raises  the  temperature 
of  the  water,  the  loss  of  heat  by  the  wet  steam  being  equal  to 
the  gain  of  heat  by  the  cold  water.  It  will,  therefore,  be  neces- 
sary to  observe  the  initial  and  final  weights,  the  initial  and 
final  temperatures  of  the  water  in  the  barrel  and  the  temperature 
of  the  steam. 

Let  W  =  original  weight  of  cold  water,  pounds. 
w  =  weight  of  wet  steam  introduced,  pounds. 
ti  =  temperature  of  cold  water,  degrees  Fahrenheit. 
/2  =  temperature  of  water  after  introducing  steam. 
ts  =  temperature  of  steam. 
L   =  latent  heat  of  steam  at  temperature  ts. 
x  =  quality  of  steam. 

Then, 

Heat  lost  by  wet  steam  =  heat  gained  by  water. 
wxL  +  w  (ts  -  fc)  =W(h-  fc) 


The  accuracy  of  the  results  depends  obviously  upon  the  accu- 
racy of  the  observation,  upon  thorough  stirring  of  the  water  so 
that  a  uniform  temperature  is  obtained  and  upon  the  length  of 
time  required.  The  time  should  be  just  long  enough  to  obtain 
accurate  differences  in  weights  and  temperatures;  otherwise, 
losses  by  radiation  will  make  the  results  much  too  low. 

Equivalent  Evaporation  and  Factor  of  Evaporation.  For  the 
comparison  of  the  total  amounts  of  heat  used  for  generating 
steam  (saturated  or  superheated)  under  unlike  conditions  it  is 
necessary  to  take  into  account  the  temperature  h  at  which  the 
water  is  put  into  the  boiler  as  well  as  also  the  pressure  P  at 
which  the  steam  is  formed.*  These  data  are  of  much  impor- 
tance in  comparing  the  results  of  steam  boiler  tests.  The  basis 

*  As  the  pressure  P  increases  the  total  heat  of  the  steam  also  increases;  but 
as  the  initial  temperature  of  the  water  ("feed  temperature")  increases  the  value 
of  the  heat  of  the  liquid  decreases. 


86  PROPERTIES  OF  VAPORS 

of  this  comparison  is  the  condition  of  water  initially  at  the  boil- 
ing point  for  "  atmospheric"  pressure  or  at  14.7  pounds  per 
square  inch;  that  is,  at  212°  F.  and  with  steaming  taking  place 
at  the  same  temperature.  For  this  standard  condition,  then, 
h  =  o  and  H  =  L  =  970.4  B.t.u.  per  pound.  Evaporation 
under  these  conditions  is  described  as, 

"  from  (a  feed-water  temperature  of)  and  at  (a  pressure  corre- 
sponding to  the  temperature  of)  212°  F." 

To  illustrate  the  application  of  a  comparison  with  this  stand- 
ard condition,  let  it  be  required  to  compare  it  with  the  amount 
of  heat  required  to  generate  steam  at  a  pressure  of  200  pounds 
per  square  inch  absolute  with  the  temperature  of  the  water  sup- 
plied (feed  water)  at  190°  F. 

For  P  =  200  pounds  per  square  inch  absolute  the  heat  of  the 
liquid  h  =  354.9  B.t.u.  per  pound  and  the  heat  of  evaporation 
(L)  is  843.2  B.t.u.  per  pound.  For  t  =  190°  F.  the  heat  of  the 
liquid  (h0)  is  157.9  B.t.u.  per  pound.  The  total  heat  actually 
required  in  generating  steam  at  these  conditions  is,  therefore, 

843.2  -f  (354.9  —  157.9)  =  1040-2  B.t.u.  per  pound. 

The  ratio  of  the  total  heat  actually  used  for  evaporation  to 
that  necessary  for  the  condition  denned  by  "  from  and  at  212°  F." 
is  called  the  factor  of  evaporation.  In  this  case  it  is  the  value 

1040.2  -T-  970.4  =  1.07. 

Calling  F  the  factor  of  evaporation,  h  and  L  respectively  the 
heats  of  the  liquid  and  of  evaporation  corresponding  to  the  steam 
pressure  and  ho  the  heat  of  the  liquid  corresponding  to  the  tem- 
perature of  feed  water,  then 


970.4 


The  actual  evaporation  of  a  boiler  (expressed  usually  in  pounds 
of  steam  per  hour)  multiplied  by  the  factor  of  evaporation  is 
called  the  equivalent  evaporation. 


PROBLEMS  87 

Vapors  as  Refrigerating  Media.  Ammonia  is  generally  used 
as  the  refrigerating  medium  in  connection  with  mechanical 
refrigeration.  Carbon  dioxide  and  sulphur  dioxide  are  also  used 
to  a  limited  extent  as  refrigerating  media. 

The  value  of  a  substance  as  a  refrigerating  medium  depends 
upon  its  latent  heat,  its  vaporization  temperature,  its  cost,  and 
upon  its  chemical  properties.  The  greater  the  latent  heat  of  a 
refrigerating  medium  the  more  heat  will  it  be  capable  of  abstract- 
ing by  evaporation.  Upon  the  vaporization  temperature  of  the 
refrigerant  at  different  pressures  depends  the  degree  of  cold  it 
can  produce  as  well  as  its  practicability  for  use  in  hot  climates 
where  low  temperature  cooling  water  cannot  be  secured. 

Goodenough's  Tables  of  Properties  of  Steam  and  Ammonia  give 
the  various  physical  properties  of  saturated,  superheated  and 
liquid  ammonia.  Peabody's  Tables  give  the  properties  of 
ammonia,  sulphur  dioxide,  carbon  dioxide  and  of  other  vapors 
used  as  refrigerating  media. 

Table  4  is  an  abridged  table  based  upon  Goodenough's  values 
for  saturated  ammonia.  Tables  5  and  6,  based  upon  Peabody's 
Tables,  show  some  of  the  properties  of  sulphur  dioxide  (S02)  and 
of  carbon  dioxide  (CC^). 

PROBLEMS 

1.  Dry  and  saturated  steam  has  a  pressure  of  100  Ibs.  per  sq.  in.  absolute. 
Calculate  the  temperature  of  the  steam,  the  volume  per  pound,  the  heat 
of  the  liquid,  the  latent  heat,  and  the  total  heat  above  32°  F.  per  pound  of 
this  steam. 

2.  Dry  and  saturated  steam  has  a  temperature  of  300°  F.    What  is 
its  pressure,  heat  of  the  liquid,  latent  heat,  and  total  heat? 

3.  A  closed  tank  contains  9  cu.  ft.  of  dry  and  saturated  steam  at  a 
pressure  of  150  Ibs.  per  sq.  in.  absolute. 

(a)  What  is  its  temperature? 

(b)  How  many  pounds  of  steam  does  the  tank  contain? 

4.  A  boiler  generates  dry  and  saturated  steam  under  a  pressure  of  200 
Ibs.  per  sq.  in.  absolute.    The  feed  water  enters  the  boiler  at  60°  F. 

(a)  What  is  the  temperature  of  the  steam? 

(b)  How  many  British  thermal  units  are  required  to  generate 

i  Ib.  of  this  steam  if  this  feed  water  is  admitted  at  32°  F.? 


88  PROPERTIES  OF  VAPORS 

(c)  How  many  British  thermal  units  are  required  to  generate 
i  Ib.  of  this  steam  from  feed  water  at  60°  F.  into  the 
steam  at  the  pressure  stated  at  the  beginning  of  this  prob- 
lem? 

5.  One  pound  of  dry  and  saturated  steam  is  at  a  pressure  of  250  Ibs. 
per  sq.  in.  absolute. 

(a)  What  is  its  internal  energy  of  evaporation? 

(b)  What  is  its  total  internal  energy  above  32°  F.? 

(c)  How  much  external  work  was  done  during  its  formation  from 

32°  F.? 

(d)  How  much  external  work  was  done  during  the' evaporation? 

(e)  How  much  external  work  was  done  during  the  change  in 

temperature  of  the  water  from  32  degrees  to  the  boiling 
point  corresponding  to  the  pressure? 

6.  Dry  and  saturated  steam  is  generated  in  a  boiler  and  has  a  tempera- 
ture of  400°  F.    The  feed  water  enters  the  boiler  at  200°  F. 

(a)  What  pressure  is  carried  in  the  boiler? 

(b)  What  is  the  total  heat  supplied  to  generate  i  Ib.  of  this  steam? 

(c)  How  much  external  work  was  done  during  its  formation? 

(d)  How  much  heat  was  used  in  increasing  the  internal  energy? 
Check  this  by   (hz  —  hi  +  IL)   noting  that  this  assumes  no 

external  work  done  in  the  heating  of  the  liquid. 

7.  One  pound  of  steam  at  a  pressure  of  100  Ibs.  per  sq.  in.  absolute  has 
a  quality  of  90  per  cent  dry.    What  is  its  temperature? 

8.  How  many  British  thermal  units  would  be  required  to  raise  the 
pound  of  steam  in  the  above  problem  from  32°  F.  to  the  boiling  point  corre- 
sponding to  the  pressure  stated? 

9.  What  would  be  the  volume  of  a  pound  of  steam  for  the  conditions 
stated  in  problem  7? 

10.  How  many  heat  units  (latent  heat)  are  required  to  evaporate  the 
steam  in  problem  7? 

11.  What  is  the  amount  of  the  total  heat  (above  32°  F.)  of  the  steam  in 
problem  7? 

12.  What  would  be  the  external  work  of  evaporation  of  the  steam  in 
problem  7? 

13.  How  much  external  work  (above  32°  F.)  is  done  in  making  steam  as 
in  problem  7? 

14.  What  is  the  internal  energy  of  evaporation  of  the  steam  in  problem  7? 

15.  What  is  the  total  internal  energy  of  the  steam  in  problem  7? 

16.  A  tank  contains  9  cu.  ft.  of  steam  at  100  Ibs.  per  sq.  in.  absolute 
pressure  which  has  a  quality  of  0.95.    How  many  pounds  of  steam  does  the 
tank  contain? 


PROBLEMS  89 

17.  Two  pounds  of  steam  have  a  volume  of  8  cu.  ft.  at  a  pressure  of  100 
Ibs.  per  sq.  in.  absolute.    What  is  the  quality? 

Calculate  its  total  heat  above  32°  F. 

1 8.  One  pound  of  steam  having  a  quality  of  0.95  has  a  temperature  of 
325°  F.    What  is  the  pressure? 

19.  One  pound  of  steam  at  a  pressure  of  225  Ibs.  per  sq.  in.  absolute  has 
a  temperature  of  441.9°  F. 

(a)  Is  it  superheated  or  saturated? 

(b)  What  is  the  total  heat  required  to  generate  such  steam  from 

water  at  32°  F.? 

(c)  What  is  its  volume? 

(d)  How  much  external  work  was  done  (above  32°  F.)  in  generat- 

ing it? 

(e)  How  much  internal  energy  above  32°  F.  does  it  contain? 

20.  One  pound  of  steam  at  a  pressure  of  300  Ibs.  per  sq.  in.  absolute  has 
a  volume  of  1.80  cu.  ft.  • 

(a)  Is  it  saturated  or  superheated? 

(b)  What  is  its  temperature? 

(c)  How  much  is  its  total  heat  above  32°  F.? 

(d)  What  is  its  total  internal  energy? 

21.  Steam  in  a  steam  pipe  has  pressure  of  110.3  Ibs.  per  sq.  in.  by  the 
gage.    A  thermometer  in  the  steam  registers  385°  F.    Atmospheric  pres- 
sure is    14  Ibs.  per  sq.  in.  absolute.    Is  the  steam  superheated,  and  if 
superheated  how  many  degrees? 

22.  Steam  at  a  pressure  of  200  Ibs.  per  sq.  in.  absolute  passes  through 
a  throttling  calorimeter.    After  expansion  into  the  calorimeter  the  tem- 
perature of  this  steam  is  250°  F.  and  the  pressure  15  Ibs.  per  sq.  in.  abso- 
lute.   What  is  the  quality? 

23.  Steam  at  a  temperature  of  325°  F.  passes  through  a  throttling 
calorimeter.    In  the  calorimeter  the  steam  has  a  pressure  of  16  Ibs.  per 
sq.  in.  absolute  and  a  temperature  of  236.3°  F.    What  is  the  quality? 

24.  Steam  at  150  Ibs.  per  sq.  in.  absolute  pressure  passes  through  a 
throttling  calorimeter.    Assuming  that  the  lowest  condition  in  the  calorim- 
eter for  measuring  the  quality  is  10°  F.  superheat  and  the  pressure  in  .the 
calorimeter  is  15  Ibs.  per  sq.  in.  absolute,  what  is  the  largest  percentage  of 
wetness  the  calorimeter  is  capable  of  measuring  under  the  above  conditions? 

25.  Prepare  a  chart  by  means  of  which  can  be  determined  the  largest 
percentages  of  wetness  a  throttling  calorimeter  will  measure  at  all  pressures 
from  50  to  300  Ibs.  per  sq.  in.  absolute. 

26.  In  a  ten-minute  test  of  a  separating  calorimeter  the  quantity  of  dry 
steam  passing  through  the  orifice  was  9  pounds.    The  quantity  of  water 
separated  was  i  pound.    What  was  the  quality? 


90  PROPERTIES  OF  VAPORS 

27.  A  barrel  contained  400  Ibs.  of  water  at  a  temperature  of  50°  F.    Into 
this  water  steam  at  a  pressure  of  125  Ibs.  per  sq.  in.  absolute  was  admitted 
until  the  temperature  of  the  water  and  condensed  steam  in  the  barrel 
reached  a  temperature  of  100°  F.    The  weight  of  the  water  in  the  barrel 
was  then  418.5  Ibs.    What  was  the  quality? 

28.  Calculate  the  factor  of  evaporation  and  the  equivalent  evaporation 
per  pound  of  coal  for  a  boiler  under  the  following  conditions:  Steam  pres- 
sure, 190  Ibs.  per  sq.  in.  gage;  feed  water  temperature,  203°  F.;  steam  appar- 
ently evaporated  per  pound  of  coal  7^  pounds;  steam  3  per  cent  wet. 

29.  Plot  on  one  chart  the  pressure-temperature  relations  of  ammonia, 
S02,  and  CO2. 

30.  Plot  on  one  chart  the  pressure-latent  heat  relations  of  ammonia, 
SO2,  andCO2. 

31.  What  conclusions  do  you  reach  from  the  graphic  results  in  problems 
29  and  30? 

32.  If  cooling  water  for  condensing  ammonia  cannot  be  secured  at  a  tem- 
perature less  than  90°  F.,  what  will  be  the  pressure  of  the  ammonia  at  the 
inlet  to  the  condenser  of  the  refrigerating  system? 


CHAPTER  VI 


ENTROPY 

Pressure-volume  diagrams  are  useful  for  determining  the 
work  (in  foot-pounds),  done  during  any  process  or  a  cycle,  but 
they  are  of  very  limited  use  in  analyz-  0, 

ing    the    heat    changes    involved.     It 
has,  therefore,  been  found  desirable  to 
make  use  of  a  diagram  which  shows    j 
directly  by  an  .  area   the   number   of    ^ 
heat  units  (instead  of  foot-pounds)  in-   j 
volved  during  any  process.     In  order    j 
that  an  area  shall  represent  this  value 
the  coordinates  must  be  such  that  their 
product  will  give  heat  units.     If  the  Entropy-  0 

ordinates  are  in  absolute  temperature,  FIG.  22.—  Analysis  of  Entropy 
the  abscissas  must  be  heat  units  per  Diagram. 


dQ 


degree  of  absolute  temperature,  that  is,     ,  for  then  T  X  ~ 


the  amount  of  heat  added  during  the  process. 

Suppose  that  the  heat  Q,  which  is  required  to  produce  a  process, 
such  as  an  expansion  or  a  compression,  is  divided  up  into  a 
number  of  small  increments  dQ  (Fig.  22)  and  that  each  small 
increment  of  heat  is  divided  by  the  average  absolute  temperature 
at  which  the  heat  change  occurs.  There  will  then  be  a  series 

of  expressions  -^  which,  when  integrated,  will  give  the  total 

change  in  the  abscissas.     This  quantity  I  —  when  multiplied  by 

the  average  absolute  temperature  between  C  and  D  will  give  the 
total  amount  of  heat  involved  during  the  process. 
Mathematically  expressed,  the  change  in  the  abscissas  is 

,.       dQ  (>dQ 

d<i>  =   -^        or  0  =:  j   -^-  (121) 


9  2  ENTROPY 

The  heat  change  involved  is 

dQ  =  Td<f>    or  Q  =    CT  d<j>.  (122) 

The  quantity  0  in  the  equations  is  known  as  the  increase  in  en- 
tropy of  the  substance,  and  may  be  denned  as  a  quantity  which, 
when  multiplied  by  the  average  absolute  temperature  occurring 
during  a  process,  will  give  the  number  of  heat  units  (in  B.t.u.) 
added  or  abstracted  as  heat  during  the  process.  The  "increase 
in  entropy"  is  employed  rather  than  entropy  itself,  because 
only  the  differences  in  entropy  are  important. 

This  definition  of  entropy  states  that  in  a  temperature-entropy 
diagram  such  as  Fig.  22,  where  the  ordinates  are  absolute  tem- 
peratures, and  the  abscissas  are  entropies  as  calculated  above 
some  standard  temperature,  the  area  under  any  line  CD  gives  the 
number  of  heat  units  added  to  the  substance  in  passing  from  a 
temperature  T\  and  entropy  <f>i  =  Oe  to  a  temperature  T2  and 
entropy  fa  =  Of  (or  the  number  of  heat  units  abstracted  in  pass- 
ing from  T2  to  Ti)  . 

Entropy  Changes  During  Constant  Pressure  Expansions  of 
Gases.  The  heat  supplied  or  abstracted  during  a  constant 
pressure  expansion  or  compression  of  a  gas  may  be  stated: 

dQ  =  wCPdt,  (123) 

or  Q  =  wCP  (T2  -  Tj).  (124) 

From  equation  (121)  the  change  in  entropy  is 


Assuming  the  specific  heat  Cp  constant,  the  change  in  entropy 
becomes  : 


(I26) 
=  wCP  (log,  T2  -  log,  Ti) 


ENTROPY  CHANGES  93 

Entropy  Changes  of  Gases  at  Constant  Volume.  The  heat 
supplied  or  abstracted  during  a  constant  volume  change  of  a  gas 
may  be  stated: 

dQ  =  wC,dt,  (128) 

or  Q  =  wC,  (T2  -  7\).  (129) 

Assuming,  as  before,  the  specific  heat  under  constant  volume 
(Cr)  constant  and  substituting  in  equation  (121),  the  change  in 
entropy  is 

0  =  wCv  I      -^  (130) 

Jfv    T 

=  wC,  (loge  T2  -  log,  Ti)  (131) 


Entropy  Changes   During  Isothermal   Processes  of  a  Gas. 

During  an  isothermal  change  the  temperature  remains  constant. 
The  heat  supplied  or  abstracted  is  consequently  equal  to  the 
heat  equivalent  of  the  work  done,  which  in  terms  of  weight  and 
temperature  is 

g  (133) 


Since  the  temperature  remains  constant,  the  change  in  entropy 
is 

gef;          •     (I34) 


=  WR  loge  y-  (l35) 

Entropy  Changes  During  Reversible  Adiabatic  Processes  of 
Gases.  The  heat  supplied  to  or  abstracted  from  an  adiabatic 
expansion  or  compression  is,  by  definition,  zero.  The  entropy 
change  in  such  a  process  is 

r2/7O 
U\J  f        ^.\ 

-£  =  °-  w6) 

T 

Hence  during  an  adiabatic  process  no  change  of  entropy  occurs. 


94  ENTROPY 

It  is  possible  for  adiabatic  processes  to  occur  which  are  not 
reversible.  These  are  processes  in  which,  not  only  no  heat  is 
added  or  abstracted,  but  a  portion  or  all  the  work  of  the  process 
may  reappear  in  the  working  medium  as  heat.  In  such  a  process 
the  entropy  does  not  remain  constant.  Reversible  adiabatic 
processes  are  sometimes  called  isoentropic  (equal  entropy)  to 
distinguish  them  from  the  irreversible  adiabatic  processes,  j 

Entropy  Changes  During  a  Carnot  Cycle.  A  pound  of  the 
working  substance  is  first  expanded  isothermally.  On  a  T-(j> 

(temperature-entropy)  diagram  (Fig. 
23),  this  process  would  be  represented 
by  line  AB,  where  the  temperature  re- 
mains constant  at  TI,  and  where  the 
entropy  increases  from  Oe  to  Of,  be- 
cause of  the  addition  of  heat  that  is  re- 
Entropy-<£  quired  to  keep  the  temperature  constant. 

FIG.  23.  —  Entropy  Diagram        The  next  process  is  adiabatic  expan- 

of  Carnot  Cycle.  „._    from   ~     ^    „        R          .g 


added  to  nor  abstracted  from  the  substance  during  this  expan- 
sion. Hence  the  entropy  remains  constant,  as  indicated  by 
BC. 

The  substance  is  now  isothermally  compressed  along  CD,  the 
temperature  remaining  constant  at  7^  and  the  entropy  decreasing 
because  of  the  abstraction  of  heat. 

The  last  process  of  the  cycle  is  the  adiabatic  compression  from 
D  to  A,  no  heat  being  added  or  abstracted,  and  the  entropy, 
therefore,  remaining  constant. 

The  heat  supplied  during  the  cycle  is  represented  by  the  area, 
ABfe  and  equals 

(fy  -  00  TV  (137) 

The  heat  abstracted  during  the  cycle  is  represented  by  the 
area,  CDef,  and  equals 

(0,  -  00  TV  (138) 

Since  the  work  of  the  cycle,  in  terms  of  heat,  equals  the  heat 
added  minus  the  heat  rejected, 


TEMPERATURE-ENTROPY  DIAGRAMS   FOR  STEAM  95 

Work  =  Area  ABfe  -  area  CDef 

=  Area  ABCD. 
In  terms  of  entropy, 

Work  =  (0,  -  <t>e)  T!  -  (0,  -  4>e)  T2 

=    (7\   -   T2)    (</>/  -  0e).  (139) 

Efficiency  of  cycle  is, 

Area 


AreaAB/e 

(T,  -  T2)  (fr  -  fr)  =  7\  -  T,  ,      , 

7\  (0/  -  0e)  Z\ 

From  the  foregoing  discussion  two  important  conclusions  may 
be  drawn  in  regard  to  the  use  of  the  T-0  diagram: 

1.  If  any  heat  process  be  represented  by  a  curve  on  a  T-<f> 
diagram,  the  heat  involved  during  the  process  is  equal  to  the 
area  under  the  curve,  that  is,  between  the  curve  and  the  axis  of 
absolute  temperature. 

2.  If  a  cycle  of   heat   processes  be  represented  on  a  T-<j> 
diagram  by  a  closed  figure,  the  net  work  done  is  equal  to  the 
enclosed  area,  that  is,  the  enclosed  area  measures  the  amount 
of  heat  that  was  converted  into  work. 

T-<f>  diagrams  are  useful  for  analyzing  heat  processes,  and  find 
particularly  useful  application  in  steam  engineering,  as  indicated 
by  the  following: 

1.  Graphical  analysis  of  heat  transfers  in  a  steam  engine 
cylinder. 

2.  Determination  of  quality  of  steam  during  adiabatic  ex- 
pansion. 

3.  Calculations  of  steam  engine  cycle  efficiencies. 

4.  Steam  turbine  calculations. 
Temperature-Entropy  Diagrams  for  Steam.    A  temperature- 

entropy  diagram  for  steam  is  shown  in  Fig.  24.  The  various 
shaded  areas  represent  the  heats  added  to  water  at  32°  F.  to 
completely  vaporize  it  at  the  pressure  PI.  The  area  ABCD  is 
the  heat  added  to  the  water  to  bring  it  to  the  temperature  of 
vaporization,  or  represents  the  heat  of  the  liquid  (h)  for  the  pres- 


96 


ENTROPY 


sure  PI.  Further  heating  produces  evaporation  at  the  constant 
temperature  Ti  corresponding  to  the  pressure  PI,  and  is  repre- 
sented by  the  area  under  the  line  CE.  When  vaporization  is 
complete,  the  latent  heat,  or  the  heat  of  vaporization  (L),  is  the 
area  DCEF.  If,  after  all  the  water  is  vaporized,  more  heat  is 
added,  the  steam  becomes  superheated,  and  the  additional  heat 
required  would  be  represented  by  an  area  to  the  right  of  E. 


Dry  Steam  Or 
turation  Lino 


o.5        1.0        a.5 

Entropy  (0) 
FIG.  24.  —  Temperature-entropy  Diagram  of  Steam. 

Calculation  of  Entropy  for  Steam.  In  order  to  lay  off  the 
increase  in  entropy  as  abscissas  in  the  heat  diagram  for  steam, 
it  is  necessary  to  determine  various  values.  For  convenience 
32°  F.  has  been  adopted  as  the  arbitrary  starting  point  for  cal- 
culating increase  of  entropy,  as  well  as  for  the  other  thermal 
properties  of  steam.  Referring  to  Fig.  25  the  entropy  of  water 
is  seen  to  be  o  at  32°  F.  In  order  to  raise  the  temperature  from 
TQ  to  TI,  h  heat  units  are  required,  and  by  the  equation  (121), 
the  entropy  of  one  pound  of  the  liquid  6  will  be 

e 

or,  assuming  the  specific  heat  of  water  to  be  unity  we  can  write 

-r? 

=  loge  Ti   -  loge  TQ   =   bge  -^  (142) 


CALCULATION  OF  ENTROPY  FOR   STEAM 


97 


In  order  to  evaporate  the  water  into  steam  at  the  boiling  point 
i,  the  latent  heat,  L,  must  be  added  at  constant  temperature 
i.  The  increase  of  entropy  during  the  "steaming"  process  is 


represented  by  —  and  is  equal  to,  in  this  case, 


Ord. 

Eahr^ 


131' 


Abs. 
Fahr. 


& 


b 


log,  T2 


JEntropy-^ 


d43) 


FIG.  25.  —  Diagram  for  Calculation  of  Entropy  of  Steam. 

Further  heating  would  produce  superheated  steam  and  the 
change  in  entropy  would  be 


/*r8uP. 
Entropy  of  superheat  =   I 

t/Jsat. 


Cvdt 


Tsat. 

=  CPlogc 


T 

•I-  sup. 


(144) 
(145) 


where  Cp  =  specific  heat  of  the  superheated  steam  at  constant 

pressure. 

Taw-  —  absolute  temperature  of  the  superheated  steam. 
TBM.   =  absolute  temperature  of  the  saturated  steam. 


98  ENTROPY 

Total  Entropy  of  steam,  which  is  the  sum  of  entropies,  corre- 
sponding to  the  various  increments  making  up  the  total  heat  of 
the  steam,  depends  upon  the  quality  of  the  steam. 

For  dry  saturated  steam  the  total  entropy  above  32°  F.  and 
temperature  T\  is  equal  to  the  sum  of  the  entropy  of  the  liquid 
and  the  entropy  of  evaporation  for  dry  saturated  steam.  This 
may  be  written 

fcat.    =   01  +  ^  (146) 

1  1 

=  log<£  +  ^-  d47) 

•I  Q  1\ 

Similarly,  for  wet  steam  whose  quality  is  x,  the  total  entropy  is 

0wet  =  0,  +  ^  (I48) 

1    1 

,       7*1      xLi 
=  loge—  +  —  -• 
^o        1\ 

For  superheated  steam  whose  final  temperature  is  rsup.  and 
the  temperature  corresponding  to  saturation  is  Ti,  the  total 
entropy  is,  by  equation  (144), 

Li   .     CT^-CPdt  ,      x 

4>suP.  =  0i  +  -=•  +   /         -fr-  (149) 

^  i      ^rsat.       L 

=   loge  p  +  |i  +  C,  bg,  I=£.  (IS0) 

-t  2          -t  1  -^  sat. 

Values  of  these  entropies  will  be  found  in  steam  tables. 

Referring  to  Fig.  25,  the  line  TIC  represents  the  increase  of 
entropy  due  to  the  latent  heat  added  during  the  steaming  process. 
If  this  steaming  process  had  stopped  at  some  point  such  that 
the  steam  was  wet,  having  a  quality  x,  this  condition  of  the  steam 
could  be  denoted  by  the  point  s,  where 

Tj 


This  relation  is  obvious,  for  a  distance  along  TIC  represents 
the  entropy  of  steam,  which  is  proportional  to  the  latent  heat 
added,  which  in  turn  is  proportional  to  the  amount  of  dry  steam 


THE  MOLLIER   CHART  99 

T  wi 
formed  from  one  pound  of  water.     In  like  manner,  -^—    is    the 


quality  of  steam  at  the  point  m  that  has  been  formed  along 
at  the  temperature  T2. 

•  It  is  thus  apparent  that  any  point  on  a  T-(j>  diagram  will  give 
full  information  in  regard  to  the  steam.  The  proportional  dis- 
tances on  a  line  drawn  through  the  given  point  between  the 
water  and  the  dry  steam  lines  and  parallel  to  the  X-axis  give 
the  quality  of  the  steam  as  shown  above.  The  ordinate  of  the 
point  gives  the  temperature  and  corresponding  pressure,  while 
its  position  relative  to  other  lines,  such  as  constant  volume  and 
constant  total  heat  lines  which  can  be  drawn  on  the  same  dia- 
gram, will  give  further  important  data. 

The  Mollier  Chart.  While  the  temperature-entropy  diagram 
is  extremely  useful  in  analyzing  various  heat  processes,  its  use 
in  representing  the  various  conditions  of  steam  is  not  so  conven- 
ient as  that  of  the  Mollier  chart.  This  chart  is  illustrated  in 
Plate  i  of  the  Appendix.  The  coordinates  consist  of  the  total 
heat  of  the  steam  above  32°  F.  and  entropy.  The  saturation 
curve  marks  the  boundary  between  the  superheated  and  satu- 
rated regions.  In  the  wet  region  lines  of  constant  volume  are 
drawn  and  in  the  superheated  regions  the  lines  of  equal  super- 
heats appear.  In  both  the  superheated  and  saturated  regions, 
lines  of  constant  pressure  are  drawn  and  the  absolute  pressures 
of  the  various  curves  are  labeled. 

By  means  of  the  Mollier  chart  (Appendix)  the  curve  for 
adiabatic  expansion  can  be  very  readily  drawn  on  a  pressure- 
volume  diagram  when  the  initial  quality  of  steam  at  cut-off 
is  known.  Assume  that  one  pound  of  wet  steam  at  a  pressure 
of  150  pounds  per  square  inch  absolute  and  quality  0.965  dry 
is  admitted  to  the  engine  cylinder  per  stroke,  and  that  there 
is  previously  in  the  clearance  space  0.2  cubic  foot  of  steam 
(see  Fig.  26),  at  exactly  the  same  condition.  The  volume  of  a 
pound  of  dry  saturated  steam  at  this  initial  pressure  of  150 
pounds  is,  from  the  steam-tables,  3.012  cubic  feet.  ,  At  0.965 
quality  it  will  be  i  (0.965  X  3.012)  =  2.905  cubic  feet.  On  the 


IOO 


ENTROPY 


scale  of  abscissas  this  amount  added  to  the  0.20  cubic  foot  in 
the  clearance  gives  3.105  cubic  feet,  the  volume  to  be  plotted  at 
cut-off.  Other  points  in  the  adiabatic  expansion  curve  can  be 
readily  plotted  after  determining  the  quality  by  the  method  given 
on  page  no. 

^,  Clearance  Steam 

Q  Cut-off 


}  Release 


0.20  3.15 

Volume  of  Steam  in  Cylinder  (Cu.  ft.) 

FIG.  26.  —  Illustrative  Indicator  Diagram  of  Engine  using  Steam  with  Expansion. 

Temperature-entropy  Diagram  for  the  Steam  Power  Plant. 
Fig.  27  illustrates  the  heat  process  going  on  when  feed  water 
is  received  in  the  boilers  of  a  power  plant  at  100°  F.,  is  heated 
and  converted  into  steam  at  a  temperature  of  400°  F.,  and  then 
loses  heat  in  doing  work.  When  the  feed  water  first  enters 
the  boiler  its  temperature  must  be  raised  from  100°  to  400°  F. 
before  any  " steaming"  begins.  The  heat  added  to  the  liquid 
is  the  area  MNCD.  This  area  represents  the  difference  between 
the  heats  of  the  liquid  (374  —  68)  or  about  306  B.t.u.  The 
horizontal  or  entropy  scale  shows  that  the  difference  in  entropy 
between  water  at  100°  and  400°  F.  is  about  0.436.* 

The  curve  NC  is  constructed  by  plotting  from  the  steam 
tables  the  values  of  the  entropy  of  the  liquid  for  a  number  of 
different  temperatures  between  100°  and  400°  F. 

If  water  at  400°  F.  is  converted  into  steam  at  that  temperature, 

*  As  actually  determined  from  Marks  and  Davis'  Steam  Tables  (pages  9  and 
15),  the  difference  in  entropy  is  0.5663  —  0.1295  or  0.4368.  Practically  it  is  im- 
possible to  construct  the  scales  in  the  figure  very  accurately. 


TEMPERATURE-ENTROPY  DIAGRAM 


101 


the  curve  representing-  the  change  is.  necessarily  a  constant  tem- 
perature line  and  therefore  a  horizontal,  CE.  Provided  the 
evaporation  has  been  complete,  the  heat  added  in  the  "  steaming  " 
process  is  the  latent  heat  or  the  heat  of  evaporation  of  steam 
(L)  at  400°  F.,  which  is  approximately  827  B.t.u. 

R 
c/ V 


M 


Supei  healed 
Steam 


G' 


0  0.130 


.566 


1.528 


Entropy— <; 
FIG.  27.  —  Temperature-entropy  Diagram  for  the  Steam  Power  Plant. 

The  change  in  entropy  during  evaporation  is,  then,  the  heat 
units  added  (827)  divided  by  the  absolute  temperature  at  which 
the  change  occurs  (400  +  460  =  860°  F.  absolute)  or 


=  0.962. 


860 


The  total  entropy  of  steam  completely  evaporated  at  400°  F. 
is,  therefore,  0.566  +  0.962,  or  1.528.*  To  represent  this  final 
condition  of  the  steam,  the  point  E  is  plotted  when  entropy 
measured  on  the  horizontal  scale  is  1.528  as  shown  in  the  figure. 
The  point  E  is  shown  located  on  another  curve  RS,  which  is 
determined  by  plotting  a  series  of  points  calculated  the  same  as 
E,  but  for  different  pressures.  The  area  MNCEF  represents  the 
total  heat  added  to  a  pound  of  feed  water  at  100°  F.,  to  produce 

*  Entropy,  like  the  total  heat  (H)  and  the  heat  of  the  liquid  (h),  is  measured 
above  the  condition  of  freezing  water  (32°  F.). 


-?  I 

•'  - 

102  ENTROPY 

steam  at  400°  F.  Area  OBCEF  represents  the  total  heat  of 
steam  (H  in  the  steam  tables)  above  32°  F.  required  to  form  one 
pound  of  steam  at  400°  F. 

If  the  steam  generated  in  the  boiler  is  now  allowed  to  expand 
adiabatically  in  an  engine  cylinder  or  in  a  turbine  nozzle  to  a 
temperature  of  100°  F.,  this  expansion  will  be  represented  by  the 
line  EG.  GN  represents  the  condensation  of  the  exhaust  steam. 
The  area  NCEG  represents  the  energy  in  the  steam  available 
for  work  in  the  steam  motor. 

PROBLEMS 

1.  One  pound  of  water  is  raised  in  temperature  from  60°  to  90°  F. 
What  is  the  increase  in  entropy? 

2.  Dry  and  saturated  steam  has  a  pressure  of  100  Ibs.  per  sq.  in.  absolute. 

(a)  Calculate  the  entropy  of  the  liquid. 

(b)  Calculate  the  entropy  of  evaporation. 

(c)  Calculate  the  total  entropy  of  the  steam. 

(d)  Check  values  in  (a),  (b),  and  (c)  with  the  steam  tables. 

3.  Steam  at  150  Ibs.  per  sq.  in.  absolute  has  a  quality  of  0.90. 

(a)  What  is  the  entropy  of  the  liquid? 

(b)  What  is  the  entropy  of  evaporation? 

(c)  What  is  the  total  entropy  of  the  steam? 

4.  Steam  having  a  temperature  of  300°  F.  has  an  entropy  of  evapora- 
tion of  1.1900.     What  is  its  quality? 

5.  Steam  having  a  pressure  of  200  Ibs.  per  sq.  in.  absolute  has  a  total 
entropy  of  1.5400. 

(a)  What  is  the  total  entropy  of  dry  and  saturated  steam  under  the 

given  pressure?   ' 

(b)  Is  the  steam  wet  or  dry? 

(c)  What  is  its  quality? 

6.  Steam  having  a  pressure  of  125  Ibs.  per  sq.  in.  absolute  is  super- 
heated 100°  F. 

(a)  What  is  the  total  entropy  of  dry  and  saturated  steam  under  the 

given  pressure? 

(b)  What  is  the  entropy  of  the  superheat? 

(c)  What  is  its  total  entropy? 

7.  Steam  having  a  pressure  of  150  Ibs.  per  sq.  in.  absolute  has  a  total 
entropy  of  1.6043. 

(a)  What  is  the  total  entropy  of  dry  and  saturated  steam  under  the 
given  pressure? 


PROBLEMS  103 

(b)  Is  the  above  steam  saturated  or  superheated?  How  can  you  tell? 

(c)  How  much  superheat  has  the  steam? 

8.  A  boiler  generates  steam  of  0.90  quality  at  a  temperature  of  350°  F. 
with  the  feed  water  admitted  at  90°  F.    What  is  the  increase  in  entropy? 

9.  Plot  a  temperature-entropy  diagram  for  one  pound  of  water  vapor 
for  the  pressures  of  15,  50, 100, 150,  200,  250.    The  diagram  should  shdw  the 
liquid  line,  the  oo  per  cent  quality  line,  the  dry  saturated  line,  and  the  50° 
and  100°  superheat  curves  for  each  of  the  given  pressures. 

~io.  What  is  the  entropy  of  steam  92  per  cent  dry  at  a  pressure  of  15  Ibs. 
per  sq.  in.  absolute? 

11.  With  a  quality  of  o.oo,  what  is  the  entropy  of  evaporation  of  steam 
at  a  pressure  of  25  Ibs.  per  sq.  in.  absolute? 

12.  What  is  the  total  entropy  of  steam  94  per  cent  dry  at  a  pressure  of 
loo  Ibs.  per  sq.  in.  absolute? 


CHAPTER  VII 
EXPANSION  AND   COMPRESSION  OF  VAPORS 

Vapors,  like  gases,  can  be  expanded  or  compressed,  but  the 
laws  governing  their  thermodynarnic  relations  are  more  complex. 
In  the  heat  changes  of  vapors  the  heat  that  is  required  to  change 
the  state  of  the  substance  must  be  accounted  for,  in  addition  to 
that  necessary  to  do  the  external  work  and  to  produce  the  change 
in  temperature  of  the  substance. 

Equation  12,  showing  the  effect  of  the  heat  added  or  abstracted 
during  an  expansion  or  a  compression,  applies  equally  to  vapors 
and  gases. 

The  external  work  (equation  12)  done  during  the  expansion  or 
compression  of  vapor  is  calculated,  as  in  the  case  of  gases,  from 
the  area  under  the  expansion  or  compression  curve.  Work, 
being  a  product  of  pressure  and  volume,  is  independent  of  the 
working  medium. 

The  internal  energy  changes  (equation  12)  involved  during  an 
expansion  or  compression  of  vapors  are  best  measured  by  dif- 
ferences. The  internal  energy  at  the  beginning  and  at  the  end 
of  an  expansion  or  compression  may  be  determined  by  reference 
to  the  vapor  tables.  The  values,  as  obtained  from  such  tables 
(Tables  3,  4,  5,  6  in  Appendix),  are  calculated  from  a  standard 
datum  temperature  of  32°  F.,  and  their  difference  gives  the  change 
in  internal  energy  involved  in  the  process  in  question. 

Thus,  for  any  vapor  the  following  general  equation  may  be 
written: 

Heat  added  =  Internal  energy  at  the  end  of  the  expansion  or 
compression  minus  the  internal  energy  at  the  beginning  of  the 
expansion  or  compression  plus  the  heat  equivalent  of  the  external 
work  done,  or 

Q  =  (h  -  /i)  +  W.  (151) 

104 


EXPANSION  OF  WET  STEAM  AT  CONSTANT  VOLUME      105 

The  following  problems  show  the  application  of  equation  (151) 
to  various  types  of  expansions. 

i.  Expansion  of  Wet  Steam  at  Constant  Volume.  Assume 
that  one  pound  of  steam  at  a  pressure  of  15  pounds  per  square 
inch  absolute,  and  50  per  cent  dry,  receives  heat  under  constant 
volume  raising  the  pressure  to  30  pounds  per  square  inch  absolute. 

Find:  (a)  the  volume  of  the  steam  after  the  addition  of  the 
heat,  (b)  the  quality  of  the  steam,  (c)  the  work  done,  (d)  the 
heat  added. 

Solution:  (a)  Since  the  volume  remains  constant  the  final 
volume  of  the  steam  is 

zFsat.  =  0.50  X  26.27  =  13.13  cu.  ft, 

in  which  x  =  the  quality  of  the  steam,  V  =  volume  of  .saturated 
steam  at  15  pounds  per  square  inch  absolute  pressure  (26.27) 
as  obtained  from  steam  tables,  in  cubic  feet  per  pound. 

(b)  From  steam  tables  the  volume  of  one  pound  of  dry  and 
saturated  steam  at  the  final  condition  of  30  pounds  per  square 
inch  absolute  is  13.74.  Since  the  actual  volume  is  less  than  that 
of  the  saturated  steam,  the  steam  in  its  final  condition  is  wet  and 
the  quality  is 

zFsat.    =    F2l 

from  which 


or 


(c)  Since  the  volume  is  constant  the  work  done  is 

W  =  o. 

(d)  The  heat  added  is 

Q  =  /2  -  /i  +  W 

=  h  -  Ii  +  o 

=   /2    -  /I. 

The  internal  energy  at  the  final  condition  is,  from  equation 
(no), 

=  A. 


2  - 

778 


106  EXPANSION  AND   COMPRESSION  OF  VAPORS 

=  (218.8  +  0.955  x  869.0) 

=  218.8  +  829.9  =  1048.7  B.t.u., 

where  fh  and  p2  are  the  heat  of  the  liquid  and  internal  latent 
heat  respectively  at  30  pounds  per  square  inch  absolute  pressure. 
The  internal  energy  at  the  initial  condition  is 

/i  =  hi  +  XIPI 

=  (181.0  +  0.50  X  896.8) 

=  (181.0  +  448.4)  =  629.4  B.t.u. 

Then, 

Q  =  1048.7  —  629.4 
=  419.3  B.t.u. 

2.   Expansion  of   Superheated   Steam  at  Constant  Volume. 

One  pound  of  steam  at  130  pounds  per  square  inch  absolute  pres- 
sure and  50°  F.  superheat  is  heated  under  constant  volume  to 
a  pressure  of  180  pounds  per  square  inch  absolute. 

Find  (a)  the  final  quality  of  the  steam; 
(b)  the  heat  supplied. 

Solution:  (a)  the  initial  volume  of  the  steam  is  found  by 
reference  to  the  superheated  steam  tables  to  be  3.74  cubic  feet. 
Since  this  equals  the  final  volume  of  the  steam,  the  final  condition 
of  the  steam  is  300°  F.  superheat,  which  from  the  steam  tables  is 
the  condition  when  steam  at  180  pounds  per  square  inch  absolute 
pressure  has  a  volume  of  3.74  cubic  feet. 

(b)  The  heat  supplied  equals 


No  values  corresponding  to  the  internal  energy  are  available 
in  the  superheated  tables  and  consequently  these  must  be  cal- 
culated. 


where 

HSU]>.  =  the  total  heat  of  superheated  steam  as  found  in  the 
superheat  tables. 


EXPANSION  AT  CONSTANT   PRESSURE  107 

P2  =  pressure  in  pounds  per  square  foot. 

F2  =  volume  of  the  superheated  steam,  cubic  feet  per  pound. 

Thus,  for  the  conditions  of  the  problem, 
/.  =  ('353-9  - 


=  I353-9  -  I24-6  =  1229.3  B.t.u. 


=  1219.7  "~  89.5  =  1130.2  B.t.u., 
from  which 

Q  =  1229.3  —  1130.2  =  99.1  B.t.u. 

3.  Expansion  at  Constant  Pressure.  One  pound  of  steam 
at  a  pressure  of  150  pounds  per  square  inch  absolute  and  a 
volume  of  1.506  cubic  feet  expands  under  constant  pressure  until 
it  becomes  dry  and  saturated. 

1.  What  is  the  quality  at  the  initial  condition? 

The  volume  of  dry  and  saturated  steam  at  the  given  pres- 
sure is  3.012  cubic  feet  per  pound. 

The  quality  then  is         I'^°    =  0.50. 
3.012 

2.  What  is  the  volume  of  the  steam  at  the  final  condition? 
The  volume  of  a  pound  is  3.012  cubic  feet  since  the  steam 

is  dry  and  saturated. 

3.  What  is  the  work  done  during  the  expansion? 

Work  =  Pi  (V2  -  Fi)  =  150  X  144  (3.012  -  1.506)  =  32,530 
foot-pounds. 

4.  How  much  heat  is  required?     Several  methods  of  reason- 
ing may  be  used  in  calculating  the  values  of  the  heat  required 
and   the   work   done   during   constant   pressure    expansion   or 
compression. 

Q  =  h  -  Ii  +  W 

1-2.  =  ha  +  Xzpi 

=  330.2  +  i  X  780.4  =  ino.6 


108  EXPANSION  AND   COMPRESSION  OF  VAPORS 

/i  =  hi  +  XIPI 

=  (330.2  +  0.50  X  780.4) 
=  720.4  B.t.u. 

The  heat  equivalent  of  the  external  work  during  the  expansion 


150  X  144  (i  X  3.012  -  0.50  X  3.012) 
=  -  -  =  41*0. 

778 

Substituting  in  equation  (151), 

Q  =  IHO.6  —  720.4  +  41.8 
=  432.0  B.t.u. 

The  amount  of  heat  required  in  the  case  of  constant  pressure 
expansion  may  be  calculated  directly: 

Q  =  (x2  -  xi)  L,  (154) 

where  x2  and  x\  equal  the  qualities  at  the  initial  and  final  con- 
ditions of  the  expansion;  .  L  equals  the  latent  heat  of  dry  and 
saturated  steam  at  the  given  pressure.  Thus,  for  the  above 
problem, 

Q  =  (i.oo  —  0.50)  863.2  =  431.6  B.t.u. 

Similarly,  the  heat  supplied  during  constant  pressure  expansion 
may  be  expressed: 

Q  =  Hi-  Hlt 

where  Hz  and  HI  are  the  total  heat  of  the  steam  above  32°  for 
the  initial  and  final  conditions  respectively.  Thus,  for  the  prob- 
lem in  question, 

Hz  =  330.2  +  i.oo  X  863.2 
HI  =  330.2  +  0.50  X  863.2 

Q    =  (330.2  +  i.oo  X  863.2)  -  (330.2  +  0.50  X  863.2) 
=  431.6  B.t.u. 

Isothermal  Lines  for  Steam.    When  the  expansion  of  steam 
occurs  at  constant  pressure  as,  for  example,  in  the  conversion 


ADIABATIC  LINES   FOR  STEAM  109 

of  water  into  steam  in  a  boiler  when  the  engines  are  working,  we 
have  isothermal  expansion.  Isothermal  lines  for  wet  steam, 
which  consists  of  a  mixture  of  water  and  its  vapor,  are,  there- 
fore, straight  lines  of  uniform  pressure.  On  a  pressure-volume 
diagram  an  isothermal  line  is  consequently  represented  by  a 
horizontal  line  parallel  to  the  axis  of  abscissas.  As  steam 
becomes  superheated  the  pressure  decreases  as  the  volume 
increases;  for  highly  superheated  vapors  the  isothermal  curve 
approaches  a  rectangular  hyperbola.  On  a  T-0  diagram,  the 
isothermal  line  is  represented  by  a  line  of  constant  temperature, 
i.e.,  by  a  line  parallel  to  the  <£-axis. 

Adiabatic  Lines  for  Steam.  Adiabatic  lines  will  have  differ- 
ent curvature  depending  upon  the  substances  used.  It  will  be 
remembered  that  the  values  of  7  are  different  for  the  various 
gases  and  therefore  the  adiabatic  line  for  each  of  these  gases 
would  have  a  different  curvature.  In  the  same  way  the  curva- 
ture of  adiabatic  lines  of  steam  will  vary  with  the  quality  of  the 
steam.  Steam  which  is  initially  dry,  if  allowed  to  expand  adia- 
batically,  will  become  wet,  the  percentage  of  moisture  which  it 
will  contain  depending  on  the  extent  to  which  the  expansion  is 
carried.  Also,  on  any  T-<j>  diagram,  an  adiabatic  (isentropic)  line 
is  represented  by  a  line  parallel  to  the  T-axis,  i.e.,  by  a  line  of 
constant  entropy.  If  steam  is  initially  wet  and  is  expanded  adia- 
batically,  it  becomes  wetter  as  a  rule.* 

In  general,  in  any  expansion,  in  order  to  keep  steam  at  the 
same  relative  dryness  as  it  was  initially,  while  it  is  doing  work, 
some  heat  must  be  supplied.  If  the  expansion  is  adiabatic  so 
that  no  heat  is  taken  in,  a  part  of  the  steam  will  be  condensed 
and  will  form  very  small  particles  of  water  suspended  in  the 
steam,  or  it  will  be  condensed  as  a  sort  of  dew  upon  the  surface 
of  the  enclosing  vessel. 

The  relation  between  pressure  and  temperature  as  indicated 
by  the  steam-tables  continues  throughout  an  expansion,  pro- 
vided the  steam  is  initially  dry  and  saturated  or  wet. 

*  When  the  percentage  of  water  in  steam  is  very  great  and  the  steam  is  ex- 
panded adiabatically  there  is  a  tendency  for  the  steam  to  become  drier.  This 
is  very  evident  from  an  inspection  of  diagrams  like  Fig.  25. 


HO  EXPANSION  AND   COMPRESSION  OF  VAPORS 

Adiabatic  Curve  for  Steam.  Whether  steam  is  initially  dry 
and  saturated  or  wet,  the  adiabatic  curve  may  be  represented 
by  the  formula:  PVn  =  constant.  The  value  of  the  index  n 
depends  on  the  initial  dryness  of  the  steam.  Zeuner  has  deter- 
mined the  following  relation: 

n  =  1.035  +  ~  (i55) 

Solving  this  when  x  =  unity  (dry  and  saturated  steam)  the 
value  of  n  is  1.135,  and  when  x  is  0.75,  n  has  the  value  i.n.* 

While  the  work  during  an  adiabatic  expansion  may  be  cal- 
culated from  the  value  of  n  as  determined  from  Zeuner's  relation 
(equation  155),  the  following  method  is  more  commonly  used. 

Since  by  definition  the  heat  added  or  abstracted  during  an 
adiabatic  process  is  zero,  the  heat  equation  becomes, 

Q  =  /•  -  A  +  W  =  o, 
or  W  =  h  -  /2  (156) 

The  work  done  by  an  adiabatic  process  can  then  be  determined 
by  the  difference  between  the  internal  energies  at  the  beginning 
and  end  of  the  process. 

In  order  to  calculate  the  internal  energies  the  qualities  must 
be  known  and  these  may  be  found  from  the  entropies. 

Quality  of  Steam  During  Adiabatic  Expansion.  Since  in  an 
adiabatic  expansion  no  heat  transfer  takes  place,  the  entropy 
remains  constant,  and,  therefore,  on  a  T-<f>  diagram,  this  condition 
is  represented  by  a  straight  vertical  line  as  cgf  (Fig.  25)  or  smk. 
Thus  for  wet  steam: 

4>  =  0  +  *|>  (i57) 

where  x  —  the  quality  or  dryness  fraction  of  the  steam,  0  = 

*  Rankine  gave  the  value  of  n  =  V,  which  obviously  from  the  results  given 
is  much  too  low  if  the  steam  is  at  all  near  the  dry  and  saturated  condition.  His 
value  would  be  about  right  for  the  condition  when  x  =  0.75.  In  an  actual  steam 
engine,  the  expansion  of  steam  has,  however,  never  a  close  approximation  to  the 
adiabatic  condition,  because  there  is  always  some  heat  being  transferred  to  and 
from  the  steam  and  the  metal  of  the  cylinder  and  piston. 


QUALITY  OF   STEAM   DURING   ADIABATIC   EXPANSION      III 

the  total  entropy  of  saturated  steam,  and  0  =  the  entropy  of 
the  water. 

Since  in  adiabatic  expansion  the  entropy  remains  constant, 
the  following  equation  can  be  written 

0i  =  <fo  (total  entropies) 
or 

4  +  *f?--«  +  *f?-  (158) 

L\  ±2 

Knowing  the  initial  conditions  of  steam,  the  quality  of  the 
steam  at  any  time  during  adiabatic  expansion  can  be  readily 
determine'd.  Thus,  suppose  the  initial  pressure  of  dry  satu- 
rated steam  to  be  100  pounds  per  square  inch  absolute,  and  the 
final  pressure  after  adiabatic  expansion  17  pounds  per  square  inch. 

From  the  steam  tables  we  find  that  the  total  entropy  0i,  for 
dry  steam  at  100  pounds  pressure,  is  1.6020;  that  is 

<f>i  =  1.6020. 

The  entropies  at  17  pounds  pressure  are  also  obtained  from  the 
tables, 

1.6020  =  0.3229  +  Xz  14215, 
whence 

#2  =  0.899. 

For  a  rapid  and  convenient  means  of  checking  the  above 
result,  the  "  Total  Heat-entropy  "  diagram  iri  the  Appendix  can 
be  used.  From  the  intersection  of  the  loo-pound  pressure  line 
and  that  of  unit  quality  ("  saturation  line")  is  dropped  a  vertical 
line  (line  of  constant  entropy  =  1.602)  to  the  1 7-pound  pressure 
line.  This  latter  intersection  is  found  to  lie  on  the  0.90  quality 
line. 

Example.  One  pound  of  steam  having  a  quality  of  0.95  at  a 
pressure  of  100  pounds  per  square  inch  absolute  expands  adia- 
batically  to  15  pounds  per  square  inch  absolute. 

i.    What  is  the  quality  at  the  final  condition? 

The  total  entropy  at  the  initial  condition  equals 

6  +  x-  =  0.4743  +  0.95  ,X  1.1277  =  1.5456. 


112  EXPANSION  AND   COMPRESSION  OF  VAPORS 

The  total  entropy  at  the  end  of  the  expansion  equals 

B  +  x  —  =  0.3133  +  14416  x. 

Since  entropy  is  constant  in  adiabatic  expansion 

°-3I33  +  1.4416*  =  1.5456, 

from  which  the  final  condition  is:  x  =  0.854. 

2.    How  much  work  is  done  during  the  expansion? 

Since  there  is  no  heat  added  the  work  done  equals  the  loss  in 
internal  energy. 

The  internal  energy  at  the  end  of  the  expansion  equals 

fe  +  Xzpz  =  181.0  +  0.854  X  896.8  =  946.9  B.t.u. 
The  internal  energy  at  the  initial  condition  equals 

hi  +  XIPI  =  298.3  +  0.95  X  806.6  =  1064.6  B.t.u. 
The  work  equals 

1064.6  —  946.9  =  117.7  B.t.u.  or  91,576  foot-pounds. 

Polytropic  Expansion  (n  =  i). 

Example.  One  pound  of  steam  has  a  pressure  of  100  pounds 
per  square  inch  absolute  and  a  quality  of  0.95.  It  expands 
along  an  n  =  i  curve  to  20  pounds  per  square  inch  absolute. 

What  is  the  quality  at  the  end  of  the  expansion? 

Solution.     The  volume  of  the  steam  at  the  initial  condition  is 

x  X  F8at. 

x  X  Fiat.  =  0.95  X  4.429  =  4.207  cubic  feet. 

Obviously,  PxFi"    =  P2F2" 

and  since  n  =  i, 

100  X  4.207  =  20  X  Vzj 

Vz  =  21.035  cubic  feet. 

The  volume  of  dry  saturated  steam  at  the  end  of  the  expan- 
sion or  at  20  pounds  per  square  inch  is  20.08.  Therefore  the 
steam  is  superheated  at  end  of  expansion.  How  is  this  known? 


GRAPHICAL  DETERMINATION  OF  QUALITY  OF  STEAM      113 

Graphical  Determination  of  Quality  of  Steam  by  Throttling 
Calorimeter  and  Total  Heat-entropy  Diagram.  It  will  be  re- 
membered that  the  throttling  calorimeter  (pages  77  to  81)  de- 
pends for  its  action  upon  the  fact  that  the  total  heat  of  steam 
which  expands  without  doing  work  remains  the  same,  the  heat  in 
excess  of  that  required  to  keep  the  steam  dry  and  saturated  going 
to  superheat  the  steam.  Suppose  that  steam  enters  the  calorim- 


1200 


rf  1    M-HTH11 
/  l^r'll  I  IIU.U4  f 


W-r  I  I  /  JAWrr/7 


K/  /V    /X/«V---/77////Wf/s 


IL50 


L60 


L70 


1.80 


Entropy 
FIG.  28.  —  Mollier  Diagram  for  Determining  Quality  of  Steam. 

eter  at  a  pressure  of  150  pounds  per  square  inch  absolute,  and  is 
throttled  down  to  1 7  pounds  per  square  inch,  the  actual  tempera- 
ture being  240°  F.  Since  the  saturation  temperature  for  steam 
at  17  pounds  pressure  is  219.4,  the  steam  in  the  calorimeter  is 
superheated  240°  —  219.4°  or  20.6  degrees.  In  order  to  find  the 
quality  of  the  live  steam  refer  to  the  "  Mollier  Diagram"  (Appen- 
dix or  Fig.  28)  and  find  the  intersection  of  the  20.6  degrees  super- 
heat line  with  the  1 7-pound  pressure  line.  From  this  point  follow 


114  EXPANSION  AND   COMPRESSION  OF  VAPORS 

a  horizontal  line  (line  of  constant  total  heat)  to  the  left  until  it 
intersects  the  i5o-pound  pressure  line.  This  point  of  intersec- 
tion is  found  to  lie  on  the  0.96  quality  line. 

Formula   117   (page  80)   gives  the  following  result  in  close 
agreement  with  the  diagram: 

Xl  = 


Q-47  (24Q  ~  219.4)  -  330.2 


863.2 
=  0.965. 

PROBLEMS 

1.  Dry  saturated  steam  at  100  Ibs.  per  sq.  in.  absolute  pressure  contained 
in  a  closed  tank  is  cooled  until  its  pressure  drops  to  15  Ibs.  per  sq.  in.  abso- 
lute.   What  is  the  final  quality  and  the  heat  removed  from  each  pound  of 
steam? 

2.  One  pound  of  steam  at  15  Ibs.  per  sq.  in.  absolute  has  a  volume  of  12.36 
cu.  ft.    It  is  heated  under  constant  volume  until  the  pressure  becomes 
50  Ibs.  per  sq.  in.  absolute,     (a)   What  is  the  quality  before  and  after  the 
heating?     (b)  How  much  heat  was  supplied? 

3.  A  closed  tank  containing  dry  and  saturated  steam  at  15  Ibs.  per  sq.  in. 
absolute  pressure  is  submerged  in  a  body  of  water  at  a  temperature  of  59°  F. 
What  will  be  the  ultimate  pressure  and  quality  of  steam  within  the  tank? 

4.  Prove  that  for  a  constant  pressure  expansion  the  heat  supplied  equals 
the  difference  between  the  total  heats  of  the  vapor  at  the  beginning  and  at 
the  end  of  the  expansion.  . 

5.  One  pound  of  steam  at  100  Ibs.  per  sq.  in.  absolute  pressure  and  50  per 
cent  dry  expands  at  constant  pressure.     What  work  is  done  and  what  heat 
is  required  to  double  the  volume?     What  is  the  temperature  at  the  beginning 
and  at  the  end  of  the  expansion? 

6.  One  pound  of  steam  at  a  pressure  of  100  Ibs.  per  sq.  in.  absolute 
and  50  per  cent  dry  is  expanded  isothermally  until  it  is  dry  and  saturated. 
Find  the  heat  supplied  and  the  work  done. 

7.  One  pound  of  steam  at  a  pressure  of  150  Ibs.  per  sq.  in.  absolute  has  a 
quality  of  0.90.    What  work  is  done  and  what  heat  is  required  to  double 
its  volume  at  constant  pressure? 

8.  If  steam  at  200  Ibs.  per  sq.  in.  absolute,  95  per  cent  dry,  is  caused  to 
expand  adiabatically  to  228°  F.,  what  are  the  properties  of  this  steam  at 
the  lower  point?     (That  is,  final  total  entropy,  entropy  of  evaporation, 
quality  and  volume.) 

9.  What  will  be  the  final  total  heat  of  dry  saturated  steam  that  is  ex- 
panded adiabatically  from  150  Ibs.  per  sq.  in.  absolute  down  to  10  Ibs. 
per  sq.  in.  absolute? 


PROBLEMS  115 

10.  Steam  having  a  quality  of  0.20  dry  is  compressed  along  an  adiabatic 
curve  from  a  pressure  of  20  Ibs.  per  sq.  in.  absolute  to  a  pressure  correspond- 
ing to  a  temperature  of  293°  F.    What  is  the  final  quality? 

11.  Determine  the  final  quality  of  the  steam  and  find  the  quantity  of 
work  performed  by  2  Ibs.  of  steam  in  expanding  adiabatically  from  250  Ibs. 
per  sq.  in.  absolute  pressure  to  100  Ibs.  per  sq.  in.  absokite,  the  steam  being 
initially  dry  and  saturated. 

12.  One  pound  of  steam  at  150  Ibs.  per  sq.  in.  absolute,  and  200°  F. 
superheat,  expands  adiabatically.    What  is  the  pressure  when  the  steam 
becomes  dry  and  saturated?    What  is  the  work  done  during  the  expan- 
sion? 

13.  One  pound  of  dry  and  saturated  steam  at  15  Ibs  per.  sq.  in.  absolute 
pressure  is  compressed  adiabatically  to  100  Ibs.  per  sq.  in.  absolute  pressure. 
What  is  the  quality  at  the  end  of  the  compression  and  the  negative  work 
done? 

14.  One  pound  of  steam  at  100  Ibs.  per  sq.  in.  absolute  pressure  has  a 
quality  of  0.80.    It  expands  along  an  n  =  i  curve  to  a  pressure  of  15  Ibs. 
per  sq.  in.  absolute,     (a)  What  is  the  volume  at  the  beginning  and  at  the 
end  of  the  expansion,  (b)  the  quality  at  the  end  of  the  expansion,  (c)  the 
work  done  during  the  expansion,  (d)  the  heat  supplied  to  produce  the  expan- 
sion? 

1 3.  One  pound  of  steam  at  150  Ibs.  per  sq.  in.  absolute  expands  along  an 
n  =  i  curve  to  15  Ibs.  per  sq.  in.  absolute.  The  quality  of  the  steam  at 
15  pounds  is  to  be  dry  and  saturated,  (a)  What  must  be  the  quality  at  the 
initial  conditions?  (b)  What  work  will  be  done  by  the  expansion?  (c)  What 
heat  must  be  supplied? 

1 6.  Steam  at  a  pressure  of  100  Ibs.  per  sq.  in.  absolute  having  a  quality 
of  0.50  expands  adiabatically  to  15  Ibs.  per  sq.  in.  absolute.    What  is  the 
quality  at  the  end  of  the  expansion? 

17.  One  pound  of  steam  at  a  pressure  of  100  Ibs.  per  sq.  in.  absolute  has 
a  volume  of  4  cu.  ft.  and  expands  adiabatically  to  15  Ibs.  per  sq.  in.  absolute. 

(a)  What  is  the  quality  at  the  initial  and  final  conditions? 

(b)  What  is  the  work  done  during  the  expansion? 

18.  One  pound  of  steam  having  a  pressure  of  125  Ibs.  per  sq.  in  abso- 
lute and  volume  of  4.17  cu.  ft.  expands  adiabatically  to  25  Ibs.- per  sq.  in. 
absolute. 

(a)  What  is  the  quality  at  the  initial  and  final  conditions? 

(b)  What  is  the  work  of  expansion? 

19.  Given  the  steam  as  stated  in  problem  18  but  with  expansion  complete 
at  100  Ibs.  per  sq.  in.  absolute.     What  is  the  quality  at  this  pressure? 

20.  What  would  be  the  pressure  if  the  steam  in  problem  18  were  expanded 
adiabatically  until  it  became  dry  and  saturated? 


Ii6  EXPANSION  AND   COMPRESSION  OF  VAPORS 

21.  One  pound  of  steam  at  a  temperature  of  360°  F.  has  a  quality  of 
0.50,  and  expands  under  constant  pressure  to  a  volume  of  3.4  cu.  ft. 

(a)  What  is  the  quality  at  the  final  condition? 

(b)  What  is  the  work  of  the  expansion? 

(c)  What  heat  is  required? 

22.  Two  pounds  of  steam  at  a  pressure  of  100  Ibs.  per  sq.  in.  absolute 
have  a  volume  of  4  cu.  ft.,  and  expand  under  constant  temperature  to  a 
volume  of  8  cu.  ft. 

(a)  What  is  the  quality  at  the  initial  and  final  conditions? 

(b)  What  is  the  work  of  the  expansion? 

(c)  How  much  heat  is  required? 


CHAPTER  VIH 
CYCLES  OF  HEAT  ENGINES  USING  VAPORS 

Carnot  Cycle.  The  Carnot  cycle  using  a  vapor  employs  the 
same  apparatus  as  was  explained  on  page  40.  The  cycle  is  made 
up  of  two  isothermals  and  two  adiabatics,  but  differs  from  the 
cycle  using  gas  as  the  working  medium  in  that  the  isothermal 
curves  are  lines  of  constant  pressure. 


Volume 
FIG.  29.  —  Carnot  Cycle  using  Vapors. 

As  an  illustration,  assume  that  the  vapor  in  the  cylinder  is  in 
the  liquid  state  and  at  a  temperature  T\  when  the  heat  is  applied. 
The  heating  process  continues  until  the  vapor  becomes  dry  and 
saturated.  In  order  to  maintain  the  temperature  constant  the 
heat  must  be  supplied  at  such  a  rate  as  to  maintain  the  pressure 
constant.  The  volume  increases  during  this  change  from  that 
of  the  specific  volume  of  the  liquid  to  the  specific  volume  of  the 
dry  saturated  vapor  at  the  temperature  T\.  This  change  is 
represented  by  the  lines  ab  on  the  P  V  and  temperature-entropy 
diagrams,  Figs.  29  and  30. 

117 


n8 


CYCLES  OF  HEAT  ENGINES  USING  VAPORS 


The  cylinder  is  then  removed  from  the  source  of  heat  and  the 
adiabatic  expansion  is  produced.  This  is  represented  in  Figs. 
29  and  30  by  the  curves  be. 

The  isothermal  compression  curve  cd  is  produced  when  the 

cylinder  is  placed  in  communi- 
cation with  the  refrigerator  or 
condenser  and  heat  is  absorbed 
from  the  vapor.  Since  the  cyl- 
inder contains  a  saturated  va- 
por, the  process  results  in  a 
constant  pressure  curve  at  tem- 
perature, T2. 

The  adiabatic  compression 
curve  da  follows  when  the  cyl- 
inder is  removed  from  the  re- 
frigerator. The  vapor  in  its 
final  condition  is  reduced  to  a 
liquid  at  the  temperature  TV 
The  heat  added  to  the  cycle 


Entropy-^ 

FIG.  30. — Temperature-entropy  Diagram 
of  Carnot  Cycle  using  Vapors. 

is  that  required  to  produce  the 
For  the  assumed  case  this  is : 


line  ab,  Figs.  29  and  30. 
ft 


dS9) 


where  HD  =  total  heat  of  dry  saturated  steam  at  the  pressure  Pb, 
ha  =  heat  of  liquid  above  32°  F.  for  condition  at  a. 

The  heat  rejected  from  the  cycle  is  that  absorbed  by  the 
refrigerator  and  for  each  pound  of  vapor  it  is  equal  to 

ft  =  (he  +  XcLc)  -  (hd  +  XaLd).  (160) 

From  the  temperature-entropy  diagram,  Fig.  30,  the  heat  added 
and  heat  rejected  are 

Q^T^fa-fa),  /       (161) 

ft  =  Tz  (<f>c  -  fa).  (162) 
Since               (<j>c  —  fa)  =  (fa  —  fa), 

ft  =  Ti  (fa  -  fa).  (163) 


THE  RANKINE   CYCLE  119 

The  work  of  the  cycle  may  be  determined  from  the  algebraic 
sums  of  the  work  under  the  individual  curves,  but  can  be  more 
simply  calculated  from  the  temperature-entropy  diagram,  Fig. 
30.  The  area  abed  is  proportional  to  the  work  and  since  it  is  a 
rectangle, 

work  of  cycle  =  (T2  —  7\)  (to  —  to)- 

From  equations  (161)  and  (163),  the  efficiency  of  the  cycle  is 
(r,  -  7\)  (to  -  to) 


(i64) 


r, 


Equation  (164)  shows  that  the  efficiency  of  Carnot's  cycle  is 
not  affected  by  the  character  of  the  working  substance  and  is 
dependent  only  upon  its  initial  and  final  temperatures. 

The  Rankine  Cycle.*  The  Carnot  cycle  gives  the  maximum 
efficiency  obtainable  for  a  heat  engine  operating  between  given 
limits  of  temperature.  In  order  that  a  steam  engine  may  work  on 
a  Carnot  cycle,  the  steam  must  be  evaporated  in  the  cylinder 
instead  of  in  a  separate  boiler,  and  condensed  in  the  cylinder 
instead  of  being  rejected  to  the  air  or  to  a  separate  condenser. 
Such  conditions  are  obviously  impracticable,  and  it  has,  therefore, 
been  found  necessary  to  adopt  some  other  cycle  which  conforms 
more  with  practical  conditions.  The  most  efficient  practical 
steam  cycle,  and  the  one  which  has  been  adopted  as  the  standard 
with  which  the  efficiency  of  all  steam  engines  may  be  compared 
is  the  Rankine  Cycle.  The  pressure- volume  diagram  of  this  cycle 
is  shown  in  Fig.  31.  Steam  is  admitted  at  constant  pressure  and 
temperature  along  ab  to  a  cylinder  without  clearance.  At  b  cut- 
off occurs,  and  the  steam  expands  adiabatically  from  b  to  c,  some 
of  i  condensing  during  the  process.  The  steam  is  then  dis- 
charged at  constant  pressure  and  temperature  along  the  back 
pressure  line  cd.  Line  da  represents  the  rise  in  temperature  and 

*  Also  known  as  the  Clausius  Cycle,  having  been  published  simultaneously 
and  independently  by  Clausius. 


I2O 


CYCLES  OF  HEAT  ENGINES  USING  VAPORS 


pressure  at  constant  volume  when  the  condensed  steam  is  con- 
verted into  a  vapor. 
The  four  stages  of  the  Rankine  cycle  may  be  stated  as  follows: 

(1)  Feed  water  raised  from  temperature  of  exhaust  to  tem- 
perature of  admission  steam.     (Line  da.) 

(2)  Evaporation  at  constant  admission  temperature.     (Line 
ab.) 

(3)  Adiabatic  expansion  down  to  back  pressure.     (Line  be.) 

(4)  Rejection  of  steam  at  the  constant  temperature  corre- 
sponding to  the  back  pressure.     (Line  cd.) 


0  Volume 

FIG.  31.  —  Indicator  Diagram  of  Ideal  Rankine  Cycle. 

The  net  work  done  in  the  cycle,  assuming  one  pound  of  wet 
steam,  is  calculated  as  follows: 


1.  External  work  of  evaporation  (in  B.t.u.), 

(Wab)  =  rhr  (PxPta). 

2.  Loss  in  internal  energy, 

(WjJ  =  h  +  XIPI  -  (fe  +  aW)  B.t.u. 


(165) 


(166) 


3.  External  work  of  evaporation  at  temperature  of  exhaust 
(B.tu.), 


Waa  =  O. 


(167) 

(l68) 


THE   RANKINE   CYCLE  121 


Adding, 
Net  work  of  cycle, 

W  =  TrsPiViXi  +  hi  +  XIPI  —  fa- 


but 

rhr-PiFi  +  Pi  =Li. 
Therefore, 

W  =  h  +  *iZi  -  (fe  +  XzLt)  (B.t.u.).  (169) 

Thus  the  net  work  of  the  Rankine  cycle  is  equal  to  the  dif- 
ference between  the  total  heat  of  the  steam  admitted  and  the 
total  heat  of  the  steam   exhausted.     This  statement  applies 
whether  the  steam  is  initially  wet,  dry  or  superheated. 
The  heat  added  during  the  cycle  is 

Qi  =  ha  +  XbLb  —  hd.  (170) 

The  heat  rejected  during  the  cycle  is 

Q2  =  xdc.  (171) 

The  efficiency  of  the  Rankine  cycle  is 


Since 

—      —  ,      ^ 

(I72) 

Fig.  32  is  the  T-<f>  diagram  for  a  Rankine  cycle  using  dry 
saturated  steam.  The  letters  abed  refer  to  the  corresponding 
points  in  the  pressure-  volume  diagram  Fig.  31.  The  net  work 
of  the  cycle  is  the  area  B  +  C,  which  is  the  difference  between 
the  total  heats  at  admission  and  exhaust.  The  heat  added  per 
cycle  is  represented  by  the  areas  A  +  B  +  C  +  D  and  the 

Thermal  efficiency  =  +  d73> 

X^LZ  ,      , 


h,  + 
*  The  final  quality  can  be  determined  by  equation  (158). 


122 


CYCLES  OF  HEAT  ENGINES  USING  VAPORS 


hz  in  the  denominator  must  always  be  subtracted  from  hi  + 
(the  total  heat  above  32°  F.),  in  order  to  give  the  total  heat  above 
the  temperature  of  feed  water,  which  in  engine  tests  is  always 
assumed  for  the  purpose  of  comparison  to  be  the  same  as  the 
exhaust  temperature. 


Entropy— 0 
FIG.  32.  —  Temperature-entropy  Diagram  of  Rankine  Cycle. 

Example.  One  pound  of  steam  at  a  pressure  of  160  pounds  per 
square  inch  absolute  and  quality  of  0.95  performs  a  Rankine  cycle 
exhausting  at  5  pounds  per  square  inch  absolute. 

1.  What  is  the  quality  of  the  exhaust? 
The  total  entropy  at  the  initial  condition 

=  0.5208  +  0.95  X  1.0431. 
The  total  entropy  at  the  exhaust 

=  0.2348  +  i. 608  x. 

Then        0.5208  +  0.95  X  1.0431  =  0.2348  +  i. 60840;. 
From  which  x  =  0.794. 

2.  What  is  the  net  work  of  the  cycle? 

Work  =  HI  —  H2  =  335.6  +  0.95  X  858.8  —  130.1  —  0.794 

X  1000.3  =  227.2  B.t.u. 

3.  What  is  the  efficiency  of  the  cycle? 


THE  RANKINE   CYCLE 


123 


Efficiency  = — *— '    n  n =0.222  or  22. 2  per  cent. 

335.6  +  0.95  X  858.8  -  130.1 

In  the  actual  steam  engine  cycle  it  is  impractical  to  expand  the 
steam  down  to  the  back-pressure  line. 

The  Rankine  cycle  for  the  actual  steam  engine  is  similar  to 
that  described,  except  that  the  adiabatic  expansion  terminates  at 
a  pressure  higher  than  that  of  the  back-pressure,  that  is,  the 
expansion  is  incomplete. 

The  pressure-volume  and  T-4>  diagrams  of  this  modified 
Rankine  cycle  using  dry  saturated  steam  are  shown  in  Figs.  33 
and  34.  The  cylinder  is  without  clearance.  Steam  is  admitted 


Volume 
FIG.  33.  —  Rankine  Cycle,  Incomplete  Expansion. 

at  constant  pressure  and  temperature  along  ab.  Cut-off  occurs 
at  point  b  and  the  steam  expands  adiabatically  to  the  terminal 
pressure  c.  Part  of  the  steam  is  discharged  at  constant  volume 
cd.  The  remainder  is  exhausted  during  the  back-pressure  stroke 
de.  Line  ea  represents  the  rise  in  temperature  of  the  feed  water 
from  T2  to  Ti. 

The  net  work  of  the  cycle  can  best  be  calculated  by  dividing 
Fig-  33  into  the  two  areas  abcc'  and  c'cde.  The  area  abcc'  is  that 
of  the  theoretical  or  ideal  Rankine  cycle,  while  c'cde  is  that  of  the 
rectangle.  The  net  work  of  the  cycle  is  equal  to  the  sum  of  these 
two  areas. 


124 


CYCLES  OF  HEAT  ENGINES  USING  VAPORS 


The  heat  equivalent  of  the  area  abcc'  is 

hi,  +  XbLt,  —  hc  —  XcLc. 
The  area  c'cde  is 

Th  (P.  ~  P«)  (V*  ~  O). 

The  total  heat  equivalent  of  the  work  of  the  cycle  is 

fc  +  xbLb  -hc-  XcLc  +  (Pc  "  Pd)  Vd. 

778 

The  heat  added  to  the  cycle  is 


—  Pd      * 


hb  +  xbLb  -  hd. 
The  efficiency  of  the  cycle  is 


J  T        i 

—  hc  —XcLc  + 


E    - 


+ 


(i75) 

(i76) 

(177) 
(178) 

d79) 


In  the  T-(j>  diagram,  Fig.  34,  the  letters  abcde  refer  to 
corresponding  points  in  the  pressure-volume  diagram  (Fig. 
33).  The  net  work  of  the  cycle  is  represented  by  the  area  B 

+  C.  The  heat  added  to  the 
cycle  is  represented  by  the  area 
A  +  B  +  C  +  D. 

The  incomplete  Rankine  cy- 
cle (Figs.  33  and  34)  is  less 
efficient  than  the  ideal  cycle 
(Figs.  31  and  32),  because  of 
failure  to  expand  the  steam 
completely.  This  loss  is  repre- 
sented in  Figs.  33  and  34  graphi- 
cally by  the  area  cdf.  This 
cycle  is  sometimes  used  as  a 
t  standard  in  preference  to  the 


Entropy  ?> 

FIG.  34.  —  Rankine  Cycle,  Incomplete 
Expansion. 


ideal  Rankine  cycle  when  the 
efficiencies  of  engines  are  com- 
pared. 

Example.     One  pound  of  steam  at  a  pressure  of  160  pounds 
per  square  inch  absolute  and  quality  of  0.95  goes  through  a 


THE  PRACTICAL  OR  ACTUAL  STEAM   ENGINE   CYCLE      125 

Rankine  cycle.  The  terminal  pressure  is  15  pounds  per  square 
inch  absolute  and  the  exhaust  pressure  5  pounds  per  square  inch 
absolute.  What  is  (a)  the  work  of  the  cycle,  (b)  the  heat  added 
to  the  cycle,  and  (c)  the  efficiency  of  the  cycle? 

Solution.    The  total  entropy  of  the  steam  at  the  initial  con- 
dition is 

0.5208  +  0.95  X  1.0431. 

The  total  entropy  of  the  steam  at  the  terminal  pressure  is 
°-3I33  +  x  ( 


Then       0.5208  +  0.95  X  1.0431  =  0.3133  +  x  (1.4416) 
from  which  the  quality  at  the  terminal  pressure  is 

x  =  0.831. 
The  volume  of  the  steam  at  the  terminal  pressure  is 

0.831  X  26.27  =  21.83. 

The  total  heat  of  the  steam  at  the  initial  condition  is 
335.6  +  0.95  X  858.8  =  1151.5. 
The  total  heat  of  the  steam  at  the  terminal  pressure  is 

181.0  +  0.831  X  969.7  =  986.8. 
The  work  of  the  cycle  is  then  (in  terms  of  heat  units), 


W  =  nsi-s  -  986.8  +  (I5  "  5)  X  2I-83  =  205.1 


The  heat  added  to  the  cycle  is 

335-6  +  0.95  x  858.8  -  130.1  =  1021.4  B.t.u. 

The  efficiency  of  the  cycle  is 

—  —  —  =  20.08  per  cent. 
1021.4 

The  Practical  or  Actual  Steam  Engine  Cycle.  In  the  steam 
engine  designed  for  practical  operation  it  is  impossible  to  expand 
the  steam  down  to  the  back-pressure  line;  and,  furthermore,  it 


126 


CYCLES  OF  HEAT  ENGINES  USING  VAPORS 


is  evident  that  some  mechanical  clearance  must  be  provided. 
The  result  is  that  in  the  indicator  diagram  from  the  actual  steam 
engine,  we  have  to  deal  with  a  clearance  volume,  and  both  incom- 
plete expansion  and  incomplete  compression  as  shown  in  Fig. 
35.  In  order  to  calculate  the  theoretical  efficiency  of  this  prac- 


i 

3 

Volume 

FIG.  35.  —  Indicator  Diagram  of  Practical  Engine  Cycle. 

tical  cycle,  it  is  necessary  to  assume  that  the  expansion  line  cd 
and  the  compression  line  fa  are  adiabatic.  Knowing  then  the 
cylinder  feed  of  steam  per  stroke  and  the  pressure  and  volume 
relations  as  determined  from  the  indicator  diagram,  one  can 
calculate  the  theoretical  thermal  efficiency  by  obtaining  the  net 
area  of  the  diagram  (expressed  in  B.t.u.)  and  dividing  by  the 
heat  supplied  per  cycle.  In  order  to  obtain  the  net  area  of  the 
diagram,  the  latter  may  be  divided  up  into  several  simple  parts 
as  follows: 

gcdi  —  area  equivalent  to  a  Rankine  cycle, 
jidej  —  a  rectangle, 

haf j  —  area  equivalent  to  a  Rankine  cycle  (negative), 
gbah  —  a  rectangle  (negative). 

The  actual  efficiency  of  the  steam  engine  is  usually  determined 
by  dividing  the  heat  equivalent  to  a  horse  power  by  the  heat  in 
the  steam  required  to  produce  a  horse  power.  Since  one  horse 


AN  ENGINE  USING  STEAM  WITHOUT  EXPANSION        127 


power  per  hour  is  equal  to 


efficiency  of  a  steam  engine  is 
E  = 


or  2545  B.t.u.,  the  actual 


2545 


WR  (H  -h}' 

In  equation  (180)  WR  is  the  water  rate  or  the  steam  consump- 
tion per  horse  power  per  hour,  H  is  the  total  heat  in  the  steam 
at  the  initial  pressure  and  quality  as  it  enters  the  engine,  h  is  the 
heat  in  the  feed  water  corresponding  to  the  exhaust  pressure. 

Efficiency  of  an  Engine  Using  Steam  Witho'ut  Expansion.  In 
the  early  history  of  the  steam  engine,  nothing  was  known  about 
the  "  expansive  "  power  of  steam.  Up  to  the  time  of  Watt  in 
all  steam  engines  the  steam  was  admitted  at  full  boiler  pressure 
at  the  beginning  of  every  stroke  and  the  steam  at  that  pressure 

carried  the  piston  forward 
to  the  end  of  the  stroke 
without  any  diminution  of 
pressure.  Under  these  cir- 
cumstances the  volume  of 
steam  used  at  each  stroke 
at  boiler  pressure  is  equal  to 
the  volume  swept  through 
by  the  piston. 

tAn  indicator  diagram  rep- 
resenting the  use  of  steam 
in  an  engine  without  expan- 
sion is  shown  in  Fig.  36. 
This  diagram  represents 
steam  being  taken  into  the  engine  cylinder  at  i  at  the  boiler 
pressure.  It  forces  the  piston  out  to  the  point  2  when  the 
exhaust  opens  and  the  pressure  drops  rapidly  from  2  to  3. 
On  the  back  stroke  from  3  to  4  steam  is  forced  out  of  the 
cylinder  into  the  exhaust  pipe.  At  4  the  pressure  rises  rapidly 
to  that  at  i  due  to  the  rapid  admission  of  fresh  steam  into 
the  cylinder.  In  this  case  the  thermal  efficiency  (E)  is  repre- 
sented by 


— *-      Volume 

FIG.  36.  —  Indicator  Diagram  of  an  Engine 
using  Steam  without  Expansion. 


128  CYCLES  OF  HEAT  ENGINES  USING  VAPORS 

E  =    work  done     =  (Pl  -  P4)  (7,  -  FQ^ 
heat  taken  in      778  (xiLi  +  hi  —  fe)' 

where  the  denominator  represents  the  amount  of  heat  taken  in, 
with  the  feedwater  at  temperature  of  exhaust,  t2.  In  actual 
practice  the  efficiency  of  engines  using  steam  without  expansion 
is  about  0.06  to  0.07,  when  the  temperature  of  condensation  is 
about  100°  F.  When  steam  is  used  in  an  engine  without  ex- 
pansion and  also  without  the  use  of  a  condenser  the  value  of 
this  efficiency  is  still  lower.  It  will  be  observed  that  under  the 
most  favorable  conditions  obtainable  the  efficiency  of  an  engine 
without  expansion  cannot  be  made  under  normal  conditions  to 
exceed  about  7  per  cent. 

In  the  actual  Newcomen  steam  engines  the  efficiency  was  very 
much  lower  than  any  of  the  values  given  because  at  every  stroke 
of  the  piston  a  very  much  larger  amount  of  steam  had  to  be 
taken  in  than  that  corresponding  to  the  volume  swept  through 
by  the  piston  on  account  of  a  considerable  quantity  of  steam 
condensing  on  the  walls  of  the  cylinder. 

Adiabatic  Expansion  and  Available  Energy.  A  practical  ex- 
ample as  to  how  the  temperature-entropy  diagram  can  be  used 
to  show  how  much  work  can  be  obtained  by  a  theoretically  per- 
fect engine  from  the  adiabatic  expansion  of  a  pound  of  steam 
will  now  be  given.  When  steam  expands  adiabatically  —  with- 
out a  gain  or  loss  of  heat  by  conduction  —  its  temperature  falls. 
Remembering  that  areas  in  the  temperature-entropy  diagram 
represent  quantities  of  heat  and  that  in  this  expansion  there  is 
no  exchange  of  heat,  it  is  obvious  that  the  area  under  a  curve  of 
adiabatic  expansion  must  be  zero ;  this  condition  can  be  satisfied 
only  by  a  vertical  line  which  is  a  line  of  constant  entropy. 

The  work  done  during  an  adiabatic  process,  while  it  cannot  be 
obtained  from  a  "heat  diagram/'  can  very  readily  be  determined 
from  the  area  under  the  adiabatic  curve  of  a  pressure-volume 
diagram,  or  better  still  by  the  use  of  steam- tables  as  follows:  In 
an  adiabatic  expansion  the  amount  of  work  done  is  the  mechani- 
cal equivalent  of  the  loss  in  internal  energy  as  explained  in  Chap- 
ter III.  Therefore,  it  is  only  necessary  to  determine  the  internal 


ADIABATIC  EXPANSION  AND  AVAILABLE  ENERGY        129 


energy  of  the  steam  at  the  beginning  and  end  of  the  adiabatic 
expansion. 

/i  -  hi  +  XIPI,  (182) 

h  =  h-z  +  %2p2.  (183) 

Work  during  adiabatic  expansion  =  loss  in  internal  energy 

W  =  (hi  +  Xipi  —  fa  —  X-2P2)  778  (foot-pounds).     (184) 

Fig.  37  is  a  temperature-entropy  diagram  representing  dry 
saturated  steam  which  is  expanded  adiabatically  from  an  ini- 
tial temperature  T\  (corresponding  to  a  pressure  PI)  to  a  lower 
final  temperature  T2  (corresponding  to  a  pressure  P2).  The 


Entropy  0a       <t>z 

FIG.  37.  —  Temperature-entropy  Diagram  for  Dry  Saturated  Steam 
Expanded  Adiabatically. 

initial  and  final  conditions  of  total  heat  (H)  and  entropy  (0) 
are  represented  by  the  same  subscripts  i  and  2.  The  available 
energy  or  the  work  that  can  be  done  by, a  perfect  engine  under 
these  conditions  is  the  area  NCEG.  It  is  now  desired  to  obtain 
a  simple  equation  expressing  this  available  energy  Ea  in  terms 
of  total  heat,  absolute  temperature  and  entropy. 

#1  =  area  OBNCEF, 

E2  =  area  OBNG'F', 

Ea  =  area  (OBNCEF  +  FGG'F')  -  OBNG'F', 

Ea  =  H!  -  H2  +  (02  -  fc)  T2.*  (185) 

*  It  should  be  observed  that  this  form  is  for  the  case  where  the  steam  is 
initially  dry  and  saturated.  For  the  case  of  superheated  steam  a  slightly  differ- 
ent form  is  required. 


130  CYCLES  OF  HEAT  ENGINES  USING  VAPORS 

An  application  of  this  equation  will  be  made  to  determine  the 
heat  energy  available  from  the  adiabatic  expansion  of  a  pound 
of  dry  saturated  steam  from  an  initial  pressure  of  165  pounds 
per  square  inch  absolute  to  a  final  pressure  of  15  pounds  per 
square  inch  absolute. 

Example.     PI  =  165       7\  =    826  degrees,  from  steam  tables. 
P2  =    I5       T2=    673.0 

^  =  1195.0  B.t.u.      "  " 

H2  =  1150.7  B.t.u.      " 


<h  =  1-7549 
Substituting  these  values  in  equation  (185),  we  have 

Ea  =  1195-0  -  US0-?  +  (i-7549  -  I-56l5)  673  =  174-5  B.t.u. 
per  pound  of  steam. 

Now  if  in  a  suitable  piece  of  apparatus  like  a  steam  turbine 
nozzle,  all  this  energy  that  is  theoretically  available  could  be 
changed  into  velocity,  then  we  have  by  the  well-known  formula 
in  mechanics,  for  unit  mass,* 

—  =  Ea  (foot-pounds)  =  Ea  (B.t.u.)  X  778, 
2£ 
V  =  V778  X  2  gEa  =  223.8  VEa,  (186) 

where  V  is  the  velocity  of  the  jet  and  g  is  the  acceleration  due  to 
gravity  (32.2),  both  in  feet  per  second. 

Solving  then  -for  the  theoretical  velocity  obtainable  from  the 
available  energy  we  obtain  the  following: 

V  =  223.8  ^174.5  =  223.8  X  13.22  =  2956  feet  per  second. 

The  important  condition  assumed  as  the  basis  for  determining 
equation  (185),  that  the  steam  is  initially  dry  and  saturated,  must 
not  be  overlooked  in  its  application.  There  are,  therefore,  two 
other  cases  to  be  considered: 

(1)  When  the  steam  is  initially  wet, 

(2)  When  the  steam  is  initially  superheated. 

*  See  Church's  Mechanics  of  Engineering,  page  672,  or  Jameson's  Applied  Me- 
chanics and  Mechanical  Engineering,  vol.  I,  page  47. 


AVAILABLE   ENERGY  OF  WET  STEAM  131 

Available  Energy  of  Wet  Steam.  The  case  of  initially  wet 
steam  is  easily  treated  in  the  same  way  as  dry  and  saturated 
steam.  Fig.  38  is  an  example  of  the  case  in  hand.  At  the  ini- 
tial pressure  PI  the  total  heat  of  a  pound  of  wet  steam  (hi  +  XiL\) 
is  represented  in  this  diagram  by  the  area  OBNCE"F".  The 
initial  quality  of  the  steam  (#1)  is  represented  by  the  ratio  of  the 

CE" 

lines  •==-•    The  available  energy  from  adiabatic  expansion  from 
CE 

the  initial  temperature  TI  (corresponding  to  the  pressure  PI)  to 

c/  TI     PI    i 


^      Entropy          <t>x       ^  £, 

FIG.  38.  —  Temperature-entropy  Diagram  of  Wet  Steam  Expanded  Adiabaticafly. 

the  final  temperature  TV  (corresponding  to  the  pressure  P2)  is  the 
area  NCE"G".  If  we  call  this  available  energy  Eaw,  we  have  by 
manipulation  of  the  areas, 

Eaw  =  area  OBNCEF  +  FGG'F'  -  OBNG'F'  -  G"E"EG, 

Eaw  =  #1  -  #2  +  (02  -  00  T2  -  (0!  -0.)  (TV  -  TV)  *    (187) 
or, 

Ea«  =  #1  -  #2  +  (02  -  00  TV  -  ^?  (i  -  *0  (TV  -  TV).  (188) 

1 1 

The  velocity  corresponding  to  this  energy  is  found  by  substi- 
tution in  equation  (186),  just  as  for  the  case  when  the  steam  was 
initially  dry  and  saturated. 


132  CYCLES  OF  HEAT  ENGINES  USING  VAPORS 

Example.  Calculations  for  the  velocity  resulting  from  adia- 
batic  expansion  for  the  same  conditions  given  in  the  preceding 
example,  except  that  the  steam  is  initially  5  per  cent  wet,  are 
given  below. 

PI  =  165  Ibs.  absolute.     TI  =    826  degrees     from  tables 
P%  —    15  Ibs.  absolute.     T2  =    673.0  degrees  " 

Hi  =  1195.0  B.t.u.    " 
H2  =  1150.7     "        " 
fc  =  1.5615 

4*  =  1.7549 

Lj.  =  856.8  B.t.u.      " 

Xi  —  i.oo  —  0.05  ==  0.95 

Eaw  =  (1195-0  -  1150-7)  +  (i-7549  -  1-5615)  673  - 


X  (i  -  0.95)  (826  -  673)  =  44.3  +  130.2  -  7.93 
Eaw  =  166.5  B.t.u.  per  pound  of  wet  steam 
V  =  223.8  VE^  =  223.8  X  Vi66.5 
=  223.8  X  12.9  =  2886  feet  per  second. 

This  result  can  be  checked  very  quickly  by  the  "total  heat- 
entropy"  or  "Mollier"  diagram  (Appendix).  The  intersection 
of  the  0.95  quality  line  and  165  pounds  pressure  line  is  found  to 
lie  on  the  1152  B.t.u.  total  heat  line.  Since  the  expansion  is 
adiabatic,  the  entropy  remains  constant.  Therefore,  following 
the  vertical  or  constant  entropy  line  (entropy  =  about  1.507) 
down  to  its  intersection  with  the  1  5  pounds  pressure  line,  we  find 
that  the  total  heat  at  the  end  of  adiabatic  expansion  is  985  B.t.u 
and 

1152  —  985  =  167  B.t.u.  available  energy,  as  above. 

If  the  steam  were  initially  superheated,  the  available  energy 
during  adiabatic  expansion  could  be  obtained  in  the  same  way  by 
means  of  the  diagram. 

Available  Energy  of  Superheated  Steam.  The  amount  of 
energy  that  becomes  available  in  the  adiabatic  expansion  of 
superheated  steam  is  very  easily  expressed  with  the  help  of  Fig. 
39.  Two  conditions  after  expansion  must  be  considered: 


AVAILABLE   ENERGY  OF  SUPERHEATED   STEAM 


133 


(1)  When  the  steam  in  the  final  condition  is  superheated, 

(2)  When  the  steam  in  the  final  condition  is  wet   (or  dry 
saturated). 

Using  Fig.  39  with  the  notation  as  before  except  Eos  is  the  avail- 
able energy  from  the  adiabatic  expansion  of  steam  initially  super- 
heated in  B.t.u.  per  pound,  <f>s  and  Hs  are  respectively  the  total 
entropy  and  the  total  heat  of  the  superheated  steam  at  the 


/ 


* 


^>HJ 


Entropy—  <£ 
FIG.  39.  —  Temperature-entropy  Diagram  for  Superheated  Srceam. 

initial  condition,  then  from  the  diagram,  when  the  steam  is 
wet  at  the  final  condition, 

Eo,  =  Hs  -  H2  +  (fc  -  4>s)  T2.  (189) 

When  the  steam  is  superheated  at  the  final  condition, 

Eos  =  Hs  -  H2'  -  (0S  -  fc')  TJ.  (190) 

It  will  be  observed  that  these  equations  (189)  and  (190)  are  the 
same  in  form  as  (184),  and  that  equation  (190)  differs  only  in  hav- 


134  CYCLES   OF  HEAT  ENGINES   USING   VAPORS 

ing  the  terms  Hs  and  <t>s  in  the  place  of  HI  and  <£i.  In  other 
words  equation  (184)  can  be  used  for  superheated  steam  if  the 
total  heat  and  entropy  are  read  from  the  steam  tables  for  the 
required  degrees  of  initial  superheat. 

The  following  examples  illustrate  the  simplicity  of  calcula- 
tions with  these  equations: 

Example  i.  Steam  at  150  pounds  per  square  inch  absolute 
pressure  and  300°  F.  superheat  is  expanded  adiabatically  to  i 
pound  per  square  inch  absolute  pressure.  How  much  energy  in 
B.t.u.  per  pound  is  made  available  for  doing  work? 

Solution.    Hs  =  1348.8  B.t.u.  per  pound, 
H2  =  1103.6    "        "     '    " 

fa    =    1.980, 
<t>s   =    1.732, 

T2  =  559.6°  F, 

Eas    =    1348.8    ~    II03.6  +    (1.980    -    1.732)   559.6 

=  383.9  B.t.u.  per  pound. 

The  result  above  may  be  checked  with  the  total  heat-entropy 
chart  (Appendix)  and  obtain  thus  (1349  —  967)  or  382  B.t.u. 
per  pound. 

Example  2.  Data  the  same  as  in  preceding  example  except 
that  the  final  pressure  is  now  35  pounds  per  square  inch  absolute. 
(Final  condition  of  steam  is  superheated.)  Calculate  Eas. 

Solution.    Hs  =  1348.8  B.t.u.  per  pound, 
HI  =  1166.8      "        "      " 


fa'  =  1.6868, 
TV  -  718.9°  F., 

Eos  =  1348.8  -  1166.8  -  (1.7320  -  1.6868)  718.9 
=  149.5  B.t.u.  per  pound. 

Application  of  Temperature-entropy  Diagram  to  Analysis  of 
Steam  Engine.  The  working  conditions  of  a  steam  engine,  as 
stated  before,  can  be  shown  not  only  by  the  indicator  card,  but 
also  by  the  employment  of  what  is  known  as  a  "  temperature- 


APPLICATION   OF   TEMPERATURE-ENTROPY   DIAGRAM      135 

entropy"  diagram.  This  diagram  represents  graphically  the 
amount  of  heat  actually  transformed  into  work,  and  in  addition 
the  distribution  of  losses,  in  the  steam  engine. 

For  illustration,  a  card  was  taken  from  a  Corliss  steam  engine 
having  a  cylinder  volume  of  1.325  cubic  feet,  with  a  clearance 
volume  of  7.74  per  cent,  or  0.103  cubic  feet;  the  weight  of  steam 
in  pounds  per  stroke  (cylinder  feed  plus  clearance)  was  0.14664 
pounds.  Barometer  registered  atmospheric  condition  as  14.5 
pounds  per  square  inch.  The  scale  of  the  indicator  spring  used 
in  getting  the  card  was  80  pounds  to  the  inch.  Steam  chest 
pressure  was  taken  as  153  pounds  per  square  inch  (absolute), 
and  a  calorimeter  determination  showed  the  steam  to  be  dry 
and  saturated. 

The  preliminary  work  in  transferring  the  indicator  card  to  a 
T-<f>  diagram,  consists  first  in  preparing  the  indicator  card  as 
follows :  It  was  divided  into  horizontal  strips  at  pressure  intervals 
of  10  pounds  with  the  absolute  zero  line  taken  as  a  reference; 
this  line  was  laid  off  14.5  pounds  below  atmospheric  conditions. 
(See  Fig.  40.)  For  reference,  the  saturation  curve  was  drawn. 
Knowing  the  weight  of  steam  consumed  per  stroke  and  the 
specific  volume  of  the  steam  (from  the  Steam  Tables),  for 
various  pressures  taken  from  the  card,  the  corresponding  actual 
volumes  could  be  obtained;  this  operation  is,  merely,  weight  of 
steam  per  stroke  multiplied  by  specific  volume  for  some  pressure 
(0.14664  X  column  5  in  the  table  below),  the  resulting  value  be- 
ing the  volume  in  cubic  feet  for  that  condition.  These  pressures 
and  volumes  were  plotted  on  the  card  and  the  points  joined,  re- 
sulting in  the  saturation  curve,  2//-6//-8//-g//. 

The  next  step  consisted  in  constructing  the  "  transformation 
table  "  with  the  columns  headed  as  shown.  All  the  condensing 
and  evaporation  processes  are  assumed  to  take  place  in  the  cylin- 
der and  the  T-<(>  diagram  is  then  worked  up  for  a  total  weight  of 
one  pound  of  steam  as  is  customary.  Column  i  shows  the  re- 
spective point  numbers  that  were  noted  on  the  card;  column  2, 
the  absolute  pressures  for  such  points;  column  3,  the  corre- 
sponding temperatures  for  such  pressures;  column  4,  the  vol- 


i36 


CYCLES  OF  HEAT  ENGINES  USING  VAPORS 


140 
120 
100 
80 


INDICATOR  CARD 


Yolume 


ENTROPY  DIAGRAM 


A 

1 

B 

l' 

5' 

1 

y 

«JKfl 

T 

A 

— 

| 

qrjn 

/ 
/ 
/ 

i 

23 

1 

3  I* 

6 
7c 

Y 

J5 
i 

R 

\ 

pen  

X 

/ 

i 

L 

20 

( 

10 
11, 

fe 

o\ 

200 

/ 

L 

is' 

*& 

^ 

1> 

^^ 

-^-x 

\ 

/ 

z 

^ 

y^ 

I 
.i_ 

IF 

150 

/ 

1" 

1615 

It 

a 

F' 

!E 

\ 

100 

/ 
/ 

\ 

\ 

/ 
-  / 

~f 

V 

A 

| 

r 

0 

M, 

M 

' 

1. 

3 

lS 

v| 

2. 

' 

Entropy 

FIG.  40.  — Temperature-entropy  Diagram  of  Actual   Steam  Engine  Indicator 

Diagram. 


APPLICATION  OF  TEMPERATURE-ENTROPY  DIAGRAM      137 


ume  in  cubic  feet  up  to  the  particular  point  measured  from  the 
reference  line  of  volumes;  column  5,  the  specific  volume  of  a 
pound  of  dry  and  saturated  steam  at  the  particular  pressure 
(Steam  Tables);  column  6,  the  volume  of  actual  steam  per 
pound,  obtained  by  dividing  the  volumes  in  column  4  by  0.14664 
pound  (total  weight  of  steam  in  cylinder  per  stroke);  column  7, 
the  dryness  fraction  "x"  found  by  dividing  column  6  by  column 

5 ;  column  8  is  the  entropy  of  evaporation  I  —  )  f or  particular  con- 
ditions (Steam  Tables);  column  9  is  the  product  of  column  7 
and  column  8;  column  10  is  the  entropy  of  the  liquid  at  various 
conditions  as  found  in  the  Steam  Tables;  column  n  is  the  sum 
of  column  9  and  column  10,  giving  the  total  entropy. 

TRANSFORMATION   TABLE 


I 

2 

3 

4 

5 

6 

7 

8 

9 

IO 

II 

umber. 

II 

iL 

"4 

«  c. 

£3s 

ill 

^TJto 

Is 

s& 

Is 

It 

S 

J1^ 

1 

"c 

§.s 

|£ 

II 

*3  C-rt 

Igg 

l| 

fl 

H 

f 

•3~ 

1 

2* 

H 

«&£* 

3^ 

Q 

c 
H 

1 

I 

145 

356 

0.1050 

3-n 

0.716 

o  .  2302 

i.  0612 

O.244O 

0.5107 

0-7547 

2 

140 

353 

0.2250 

3-22 

1.536 

0.4770 

1-0675 

0.5090 

0.5072 

i  .0162 

5 

120 

0.4230 

3-73 

2.890 

0.7740 

1-0954 

0.8475 

0.4919 

!-3394 

9 

48 

279 

0.8500 

8.84 

5.810 

0.6580 

1-2536 

0.8250 

0.4077 

1.2327 

12 

2O 

228 

1.4225 

20.08 

9.700 

0.4830 

1-3965 

0.6740 

0-3355 

1.0095 

IS 

8 

183 

0.9758 

47-27 

6.660 

O.I4IO 

0.2168 

0.2673 

0.4841 

18 

8 

183 

0.3500 

47.27 

2.390 

0.0506 

1.538010.0778 

0.2637 

0-3451 

20 

30 

250 

0.1825 

13-74 

1.245 

0.0907 

i  .3311  0.1208 

0.3680 

0.4888 

Above  table  is  employed  for  transferring  the  P-V  diagram  to  the  T-<f>  diagram. 

After  this  table  was  completed,  columns  3  and  1 1  were  plotted 
(Fig.  40).  Convenient  scales  were  selected,  the  ordinates  as  tem- 
peratures and  the  abscissas  as  entropies.  The  various  points, 
properly  designated,  were  connected  as  shown  on  the  T-<f>  dia- 
gram, the  closed  diagram  resulting.  This  area  shows  the  amount 
of  heat  actually  transformed  into  work.  This  diagram  is  the 
actual  temperature-entropy  diagram  for  the  card  taken  and  may 


138  CYCLES  OF  HEAT  ENGINES  USING  VAPORS 

be  superimposed  upon  the  Rankine  cycle  diagram  in  order  to 
determine  the  amount  and  distribution  of  heat  losses. 

The  water  line,  A- A' ',  and  the  dry  steam  line,  C-C',  were 
drawn  directly  by  the  aid  of  Steam  Table  data,  i.e.,  the  entropy 
of  the  liquid  and  the  entropy  of  the  steam  taken  at  various 
temperatures,  and  plotted  accordingly. 

Before  the  figure  could  be  studied  to  any  extent,  the  theoreti- 
cal (Rankine)  diagram  had  to  be  plotted,  assuming  that  the  steam 
reaches  cut-off  under  steam-chest  conditions;  that  it  then  ex- 
pands adiabatically  down  to  back  pressure  and  finally  exhausts 
at  constant  pressure  to  the  end  of  the  stroke  without  compres- 
sion. This  diagram  is  marked,  A-C-E-H,  on  the  T-<j>  plane. 
The  steam-chest  pressure  of  153  pounds  per  square  inch  absolute 
fixes  the  point,  C,  when  the  temperature  line  cuts  the  steam  line, 
C-C' .  The  rest  of  the  cycle  is  self-evident. 

Referring  to  the  T-<f>  diagram,  Fig.  40,  H-A-C-E  is  the  Ran- 
kine cycle  with  no  clearance  for  one  pound  of  working  fluid. 
The  amount  of  heat  supplied  is  shown  by  the  area,  Mi-H-A-C-N, 
and  of  this  quantity,  the  area  Mi-H-E-N  *  would  be  lost  in 
the  exhaust  while  the  remainder,  H-A-C-E,  would  go  into 
work.  This  is  theoretical,  but  in  practice  there  are  losses,  and 
for  that  reason,  the  Rankine  cycle  is  used  merely  for  comparison 
with  the  actual  card  as  taken  from  a  test.  The  enclosed  irregu- 
lar area,  1-2-3  •  •  •  22~23>  is  the  amount  of  heat  going  into 
actual  work.  By  observation,  it  is  evident  that  a  big  area  re- 
mains; this  must  represent  losses  of  some  sort  or  other.  That 
quantity  of  work  represented  by  the  area,  i-$-$'-i',  is  lost  on 
account  of  wire-drawing;  the  area  $'-C-D-F  shows  a  loss  due 
to  initial  condensation;  the  loss  due  to  early  release  is  shown 
by  the  area  F-12-i^-F'  for  the  real  card,  and  by  D-G-E  for  the 
modified  Rankine  cycle  (such  a  loss,  in  other  words,  is  due  to 
incomplete  expansion);  that  quantity  represented  by  22-B-i'-i 
is  lost  on  account  of  incomplete  compression,  and  H-A-B-iS  is 

*  The  areas  Mi  HACN  and  Mi  HEN  should  have  added  to  them,  the  rectangu- 
lar area  between  MiN,  and  the  absolute  zero  line,  which  is  not  shown  on  the 
diagram  (Fig.  40). 


APPLICATION   OF  TEMPERATURE-ENTROPY  DIAGRAM      139 


800 


700 


600 


500 
A' 


400 


300 


100 


Thermal  Eft.  =— -, 


Fahr.  Line 


W 


M,HA 


N 


\ 


.8  12 

Entropy 


16 


20 


FIG.  41.  —  Temperature  (absolute)  -entropy  Diagram  of  Actual  Steam  Engine 

Indicator  Diagram. 


140  CYCLES  OF  HEAT  ENGINES  USING  VAPORS 

the  loss  due  to  clearance.  The  expansion  line  from  5  on  to  9  in- 
dicates that  there  is  a  loss  of  heat  to  the  cylinder  walls,  causing 
a  decrease  of  entropy;  from  9  on  to  10,  re-evaporation  is  taking 
place  (showing  a  gain  of  entropy). 

All  of  the  heat  losses  are  not  necessarily  due  to  the  transfer 
of  heat  to  or  from  the  steam,  as  there  may  be  some  loss  of  steam 
due  to  leakage.  In  general,  however,  the  T-<j>  diagram  is  satis- 
factory in  showing  heat  losses. 

Fig.  41  was  constructed  for  the  purpose  of  showing  how  the 
actual  thermal  efficiency  and  the  theoretical  thermal  efficiency 
(based  on  the  Rankine  cycle)  can  be  obtained  from  the  T-<j> 
diagram.  The  letters  in  Fig.  40  refer  to  the  same  points  as 
in  Fig.  41,  the  only  difference  between  the  two  diagrams  being 
the  addition  of  the  absolute  zero  temperature  line  to  Fig.  41. 

rp,          i    re  •  Work  done 

Thermal  efficiency  =  —  —       .  .     • 
Heat  added 

A  <     i  -LI.         i    cc.  •  Shaded  area  W 

Actual  thermal  efficiency  =    ,,/TT  --  „  _T/> 

MI  H—A—C—J\ 

in"  (Planimeter) 


26.00  sq.  in. 
=  0.096,  or  9.6  per  cent.* 

Rankine  cycle  efficiency  (theoretical  thermal  efficiency) 

H-A-C-E 


O/~\ 

=    ~,'  v  (by  planimeter) 
26.00 

=  0.203,  or  20.3  per  cent. 

Combined  Indicator  Card  of  Compound  Engine.  The  method 
of  constructing  indicator  diagrams  to  a  common  scale  of  vol- 

*  This  value  may  be  regarded  as  the  actual  thermal  efficiency  for  one  stroke. 
inasmuch  as  in  the  succeeding  strokes  the  "cushion"  steam  will  be  used  over  and 
over  again  and  hence  will  not  constitute  a  heat  loss.  The  actual  thermal  efficiency 

W 
of  a  steam  engine  under  running  conditions  would  then  be  given  by      , 

where  18  BCE  is  a  Rankine  cycle  with  clearance  and  complete  compression. 


COMBINED  INDICATOR   CARD   OF  COMPOUND   ENGINE      141 

ume  and  pressure  shows  where  the  losses  peculiar  to  a  compound 
steam  engine  occur,  and,  to  the  same  scale,  the  relative  work 
areas. 

As  the  first  step  divide  the  length  of  the  original  indicator  dia- 
grams into  any  number  of  equal  parts  (Fig.  42),  erecting  per- 
pendiculars at  the  points  of  division. 

In  constructing  a  combined  card,  select  a  scale  of  absolute 
pressure  for  the  ordinates  and  a  scale  of  volumes  in  cubic  feet 


145.6* 


43.S6* 


!  6.76*; 


Zero  Line 


FIG.  42.  —  High  and  Low  Pressure  Indicator  Diagram  of  Compound 
Steam  Engine. 

for  the  abscissas.     To  the.  scale  adopted  draw  in  the  atmos- 
pheric pressure  ("at.")  line,  see  Fig.  43. 

Lay  off  the  low-pressure  clearance  volume  on  the  #-axis  to 
the  scale  selected.  In  like  manner  lay  off  the  piston  displace- 
ment of  the  low-pressure  cylinder  and  divide  this  length  into  the 
same  number  of  equal  parts  as  the  original  indicator  diagrams 
were  divided.  From  the  original  low-pressure  card  (Fig.  42),  de- 
termine the  pressures  at  the  points  of  intersection  of  the  perpen- 
diculars erected  above  the  line  of  zero  pressure,  taking  care 
that  the  proper  indicator-spring  scale  is  used.  Lay  off  these 
pressures  along  the  ordinates  (Fig.  43),  connect  the  points  and 
the  result  will  be  the  low-pressure  diagram  transferred  to  the 
new  volume  and  pressure  scales. 


142  CYCLES  OF  HEAT  ENGINES  USING  VAPORS 

The  high-pressure  diagram  is  transferred  to  the  new  volume 
and  pressure  scale  by  exactly  the  same  means  as  described  for 
the  low-pressure  diagram. 

The  saturation  curve  is  next  drawn.  This  curve  represents 
the  curve  of  expansion  which  would  be  obtained  if  all  the  steam 
in  the  cylinder  was  dry  and  saturated.  It  is  very  probable 
that  the  saturation  curve  for  each  cylinder  will  not  be  continuous 
since  the  weight  of  the  cushion  steam  in  the  low-pressure  cylin- 
der is  usually  not  the  same  as  that  in  the  high-pressure  cylinder. 
The  saturation  curve  would  be  continuous  for  the  two  cards  only 
when  the  weight  of  cushion  steam  in  the  high-pressure  and  low- 
pressure  cylinders  is  the  same  (assuming  no  leakage  or  other 
losses). 

On  the  assumption  that  the  steam  caught  in  the  clearance 
spaces  at  the  beginning  of  compression  is  dry  and  saturated,  the 
weight  of  the  cushion  steam  can  be  calculated  from  the  pressure 
at  the  beginning  or  end  of  compression,  the  corresponding  cylin- 
der volumes  and  the  specific  volumes  corresponding  to  the  pres- 
sure (as  obtained  from  steam- tables). 

The  total  weight  of  steam  in  the  cylinder  is  the  weight  of  steam 
taken  into  the  engine  per  stroke  plus  the  weight  of  steam  caught 
in  the  clearance  space  (cushion  steam).  The  saturation  curves 
may  now  be  drawn  by  plotting  the  volumes  which  the  total 
weight  of  steam  will  occupy  at  different  pressures,  assuming  it 
to  be  dry  throughout  the  stroke. 

The  quality  curve  (Fig.  43)  shows  the  condition  of  the  steam 
,as  the  expansion  goes  on.  At  any  given  absolute  pressure,  the 
volume  up  to  the  expansion  line  shows  the  volume  of  the  wet 
vapor,  while  the  volume  up  to  the  saturation  curve  shows  the 
volume  that  the  weight  of  the  wet  vapor  would  have  if  it  were 
dry.  Thus  at  any  given  absolute  pressure,  the  ratio  of  the  vol- 
ume of  the  wet  vapor  (as  given  by  the  expansion  line  of  the 
indicator  card)  to  the  total  volume  of  the  dry  vapor  (as  obtained 
from  the  saturation  curve)  is  the  measure  of  the  quality  of  the 
steam. 

Showing  the  quality  by  the  use  of  the  figure,  we  have  Vol. 


COMBINED  INDICATOR   CARD   OF   COMPOUND   ENGINE      143 


160 
155 
150 
145 
140 
135 
130 
125 
120 
115 
110 
105 
100 


Hj.P.  Saturatioi 


Cu 


b    d 


3  75 

£  70 
865 


,5  1  1.5  2  2.5  3  3.5  4  i.5 

Volume- Cu.  Ft, 

FIG.  43.  —  Combined  Indicator  Diagrams  for  Compound  Engine. 


144  CYCLES  OF  HEAT  ENGINES  USING  VAPORS 

ab  -i-  Vol.  ac  =  quality.  By  laying  off  this  ratio  from  a  hori- 
zontal line  to  any  scale  desired  as  shown,  the  quality  curve  may 
be  constructed. 

Hirn's  Analysis.  In  the  study  of  steam-engine  performance 
the  action  of  the  steam  in  and  the  influence  of  the  walls  of  the 
cylinder  become  a  matter  of  considerable  importance.  The 
amount  of  heat  lost,  restored  and  transformed  into  work  as  a 
result  of  variation  in  the  condition  of  the  steam  throughout  the 
engine-cycle  can  be  determined  either  by  means  of  the  entropy- 
temperature  diagram,  or  by  calorimetric  method  based  on  Hirn's 
theory. 

The  calorimetric  method,  or  more  popularly  called  Hirn':> 
Analysis,  was  developed  by  Professor  V.  Develshauvers-Devy  of 
Liege.  (See  Table  of  the  Properties  of  Steam  by  V.  Devels- 
hauvers-Devy, Trans.  Am.  Soc.  Mech.  Eng.,  Vol.  XI;  also  Pea- 
body's  Thermodynamics.) 

In  order  to  apply  Hirn's  Analysis  to  engine  testing  besides  the 
usual  readings  the  following  items  must  be  determined: 

1 .  Absolute  pressures  at  cut-off,  release  and  compression  from 
indicator  cards, 

2.  Per  cent  of  stroke  at  cut-off,  release  and  compression  from 
indicator  cards, 

3.  Weight  of  steam  per  stroke  determined  by  weighing  the 
condensed  steam  or  boiler  feed  water  and  computing  the  steam 
used  per  stroke, 

4.  Weight  of  the  cooling  water  and  its  temperature  at  inlet 
and  outlet  from  condenser, 

5.  Temperature  of  the  condensed  steam. 

Let  w  =  pounds  of  steam  supplied  to  the  cylinder  per  stroke 
at  pressure  p  and  quality  x. 

Then  the  amount  of  heat  brought  in  by  the  steam  into  the 
cylinder  per  stroke, 

Q  =  w(h  +  xL\  (191) 

where  L  =  heat  of  vaporization  and  h  =  heat  of  the  liquid 
above  the  freezing  point. 


HIRN'S  ANALYSIS 


145 


At  the  end  of  compression  a  certain  amount  of  steam  is  left 
in  the  clearance  space  and  this  steam  mingles  with  the  w  pounds 
of  steam  at  admission.  Calling  the  weight  of  the  steam  caught 
in  the  clearance  space  wi}  the  total  amount  of  steam  at  admission 
is  w  +  w\. 

According  to  thermodynamic  laws  the  addition  of  heat  to  a 
substance  produces  in  that  substance  internal  and  external 
changes,  the  internal  changes  being  volume  and  temperature 
variations  which  are  called  intrinsic  energy  changes,  while  the 


Volume 
FIG.  44.  —  The  Steam  Engine  Cycle. 

external  changes  are  changes  in  external  potential  energy  or  work 
done.  Thus  the  steam  brought  into  the  cylinder  per  stroke  on 
account  of  its  available  heat  produces  intrinsic  energy  changes 
and  is  capable  of  doing  external  work  in  overcoming  resistance. 
On  the  other  hand  the  steam  w\  caught  in  the  clearance  space  at 
compression  is  able  to  produce  only  internal  or  intrinsic  energy 
changes.  The  intrinsic  energy  changes  can  be  calculated  knowing 
the  heat  equivalent  of  internal  work  or  internal  latent  heat  (p) 
and  the  heat  of  the  liquid  (ti).  The  external  work  can  be  deter- 
mined from  the  indicator  cards. 

Referring  to  Fig.  44  and  calling  the  heat  absorbed  by  the 
cylinder  walls  during  admission  and  expansion  Qa  and  Qb\  that 
restored  during  exhaust  and  compression  Qc  and  Qd',  the  intrinsic 


146  CYCLES   OF  HEAT  ENGINES  USING  VAPORS 

energy  at  the  points  of  admission,  cut-off,  release,  and  compres- 
sion /i,  /2,  /3,  A;  the  external  work  in  heat  units  AWa,  AWb, 
AWC,  AWdj  during  the  events  of  the  stroke,  where  A  =  y^; 
also  if  the  temperature  of  the  condensed  steam  is  ts  and  that  of 
the  cooling  water  U  and  t0  at  inlet  and  outlet,  the  total  weight  of 
cooling  water  used  being  G,  we  have  the  following  relations: 

Qa   =   Q  +  h   ~  h   ~  AWa.  (192) 

ft    =/»-/.-  AW*.  (I93) 

Qc=h-h-  whs    -  G  (ho  -  fa]  -  AWc.        (194) 
Qd  =  1*-^  +  AWd.  (195) 

In  equations  (192),  (193),  (194),  and  (195)  Q  can  be  determined 
from  Q  =  w  (h  +  xL).  The  values  of  G  can  be  determined  by 
weighing  cooling  water;  h0,  hi,  and  hs  by  taking  the  temperature 
to,  tij  and  ts  by  thermometers  and  finding  h0,  hi,  and  hs  from  steam 
tables.  The  intrinsic  energy  at  the  events  of  the  stroke  are  calcu- 
lated by  means  of  the  following  equations: 

/i  =  Wi  (hi  +  XIPI),  (196) 


/a  =  (MI  +  w)  (fe  +  tfsps),  (198) 

74  =  u>i  (h^  +  #4p4),  (199) 

in  which 

Wi  =  weight  of  steam  caught  in  clearance  space. 

w   =  weight  of  steam  brought  into  the  cylinder  per 

stroke. 

hi,  h2,  h3,  hi  =  heat  of  the  liquid  at  events  of  stroke. 
Pi,  P2,  PS,  P4   =  internal  latent  heats  at  the  events  of  stroke. 
Xij  %2,  #3,  #4  =  quality  of  steam  at  events  of  stroke. 

From  the  above  it  is  evident  that  x\y  #2,  #3,  #4,  and  w  are  un- 
known, these  values  being  determined  by  the  following  procedure  : 
The  volume  of  w  pounds  of  steam  in  cubic  feet  is 

V  =  w  (xu  +  or).  (200) 

x  =  quality  of  steam,  u  =  the  increase  in  volume  produced  by 
the  vaporization  of  one  pound  of  water  of  volume  a  into  dry 


CLAYTON'S  ANALYSIS  OF   CYLINDER  PERFORMANCE      147 

steam  of  volume  5,  or  u  =  5  —  cr.  Calling  the  volume  of  the 
steam  caught  in  the  clearance  space  Vi  and  that  at  cut-off,  release 
and  compression  F2,  F3,  F4, 

Vi  =  Wi  (XiUi  +  a).  (20l) 

Fi  +  F2  =  (w  +  Wi)  (XM  +  er).  (202) 

Vi  +  V3  =  (w  +  Wi)  (X3U3  +  <r).  (203) 

Vi  +  F4  =  wi  (#4«4  -f  o-).  (204) 

In  the  above  equations  the  absolute  pressures  and  volumes  can 
be  determined  from  the  indicator  cards,  provided  the  dimensions 
of  the  engine  and  the  clearance  are  known  and  it  is  assumed 
x*  is  unity.  This  assumption  can  be  made  without  much  error, 
as  the  steam  at  compression  is  very  nearly  dry.  If  Xt  is  assumed 
equal  to  i,  then,  from  equation  (204), 

wi  =  ~J—  (205) 

«i  +  a 

The  intrinsic  energy  at  the  events  of  the  stroke  is  computed 
after  finding  Xi,  Xz,  and  x3  from  equations  (201),  (202),  and  (203) 
by  the  aid  of  equation  (205).* 

Clayton's  Analysis  of  Cylinder  Performance.  The  indicator 
diagram  has  been  used  in  determining  the  economy  of  the  steam 
engine  although  inaccuracies  enter  because  of  cylinder  condensa- 
tion. The  quantity  of  steam  admitted  to  the  engine  per  cycle 
can  theoretically  be  determined  from  the  difference  between  the 
weight  of  steam  accounted  for  at  the  point  of  cut-off  and  point  of 
compression.  Little  error  is  introduced  in  determining  the 
weight  of  steam  present  in  the  cylinder  at  the  point  of  compres- 
sion, for  the  quality  of  the  steam  at  that  point  can  be  fairly  accu- 
rately estimated.  To  determine  the  weight  of  steam  present  at 
the  point  of  cut-off  the  quality  of  the  steam  must  be  known.  The 
quality  of  the  steam  at  the  point  of  cut-off  is  not  the  same  as  that 
at  admission  because  of  the  condensation  within  the  cylinder. 
Thus  the  study  of  cylinder  performance  from  the  indicator  dia- 
gram is  made  inaccurate. 

*  For  log  form  of  Hirn's  Analysis  see  Experimental  Engineering  by  Carpenter 
and  Diederichs,  Seventh  Edition,  pages  799-806. 


148  CYCLES   OF  HEAT  ENGINES   USING  VAPORS 

Clayton's  analysis  was  made  to  determine  the  relation  be- 
tween the  quality  of  the  steam  at  cut-off  and  other  variables. 
From  the  results  of  a  careful  study  of  the  forms  of  the  expansion 
and  compression  curves  which  occur  in  indicator  diagrams,  it  has 
been  discovered  that  the  value  of  n  for  the  expansion  curve  bears 
a  definite  relation  to  the  proportion  of  the  total  weight  of  steam 
mixture  which  was  present  at  the  point  of  cut-off. 

To  facilitate  the  analysis,  the  indicator  diagram  is  transferred 
to  logarithmic  coordinates.  The  equation  of  the  poly  tropic 
curve,  PVn  —  C,  becomes  a  straight  line  when  plotted  upon 
logarithmic  cross-section  paper,  the  value  of  n  becoming  the  slope 
of  the  curve.  From  the  new  diagram,  the  slope  of  the  expansion 
and  compression  curves  may  be  easily  determined.  The  quality 
of  the  steam  at  cut-off  may  then  be  estimated,  knowing  the  speed 
of  the  engine,  the  quality  of  the  steam  at  admission,  the  pressure 
of  the  steam  and  other  values  that  affect  the  slope  of  the  curve.* 

PROBLEMS 

1.  Assume  that  i  Ib.  of  steam  of  a  pressure  of  160  Ibs.  per  sq.  in.  abso- 
lute and  a  quality  of  0.95  performs  an  ideal  Rankine  cycle,  being  exhausted 
at  a  pressure  of  5  Ibs.  per  sq.  in.  absolute.     Compute  the  quality  of  the  steam 
exhausted,  the  efficiency  of  the  cycle,  and  the  final  volume. 

2.  What  is  the  work  of  an  ideal  Rankine  cycle  if  the  steam  initially  at 
200  Ibs.  per  sq.  in.  absolute  pressure,  superheated  200°  F.,  goes  through  such 
a  cycle  with  a  back  pressure  of  i  Ib.  per  sq.  in.  absolute? 

3.  One  pound  of  steam  at  a  pressure  of  100  Ibs.  per  sq.  in.  absolute 
with  a  quality  of  0.90  performs  an  ideal  Rankine  cycle  exhausting  at  a  back 
pressure  of  2  Ibs.  per  sq.  in.  absolute.     What  is  the  net  work  and  efficiency 
of  the  cycle? 

4.  Two  pounds  of  steam  at  a  pressure  of  125  Ibs.  per  sq.  in.  absolute 
and  a  volume  of  8.34  cu.  ft.  perform  an  ideal  Rankine  cycle.     The  exhaust 
pressure  is  25  Ibs.  per  sq.  in.  absolute.    What  is  the  net  work  and  efficiency 
of  the  cycle? 

5.  One  pound  of  steam  at  a  pressure  of  160  Ibs.  per  sq.  in.  absolute  and 
a  quality  of  0,95  passes  through  a  modified  Rankine  cycle.     The  terminal 

*  For  details  concerning  Clayton's  "Analysis  of  the  Cylinder  Performance 
of  Reciprocating  Engines,"  see  Bulletin  No.  58  of  the  Engineering  Experiment 
Station  of  the  University  of  Illinois. 


PROBLEMS  149 

pressure  is  u  Ibs.  per  sq.  in.  absolute  and  the  exhaust  pressure  5  Ibs.  per  sq. 
in.  absolute.    What  is  the  efficiency  of  the  cycle? 

6.  One  pound  of  steam  at  a  pressure  of  200  Ibs.  per  sq.  in.  absolute  and 
200°  F.  superheat  passes  through  a  modified  Rankine  cycle.    The  terminal 
pressure  is  15  Ibs.  per  sq.  in.  absolute  and  the  exhaust  pressure  10  Ibs.  per 
sq.  in.  absolute.    What  is  the  net  work  and  efficiency  of  the  cycle? 

7.  Two  pounds  of  steam  at  a  pressure  of  125  Ibs.  per  sq.  in.  absolute  and 
volume  of  8.34  cu.  ft.  pass  through  a  modified  Rankine  cycle.    The  terminal 
pressure  is  20  Ibs.  per  sq.  in.  absolute  and  the  exhaust  pressure  15  Ibs.  per 
sq.  in.  absolute.    What  is  the  efficiency,  net  work,  and  heat  added  to  the 
cycle? 

8.  A  non-condensing  steam  engine  receives  steam  of  0.95  quality  and 
zoo  Ibs.  per  sq.  in.  absolute  pressure.    The  exhaust  pressure  is  15  Ibs.  per 
sq.  in.  absolute  and  the  steam  consumption  of  the  engine  is  30  Ibs.  per 
indicated  horse  power  per  hour.     What  is  the  thermal  efficiency  of  the 
engine?    What  is  the  efficiency  of  the  ideal  Rankine  cycle  operating  under 
the  same  conditions?    What  is  the  efficiency  of  the  Carnot  cycle  operating 
between  the  same  temperature  limits? 


Throat  or  smallest 
section  of  nozzle 


CHAPTER  IX 
FLOW  OF  FLUIDS 

Flow  through  a  Nozzle  or  Orifice.    Thermodynamic  prob- 
lems embrace  the  measurement  of  the  flow  of  air  or  of  a  mix- 
ture of  a  liquid  and  vapor  through  a  nozzle  or  orifice.     In  the 
A  B     nozzle  shown  in  Fig.  45,  let  A  and  B 

be  two  sections  through  which  the  sub- 
stance passes.    At  A  let  a  pressure  of 
PI  be  maintained  and  at  B  a  pressure 
12   of  P%.    To  maintain  the  constant  pres- 
sure  at  A  of  PI  let  more  substance  be 
FIG.  45.  —  Typical  Nozzle  added,  while  at  B  allow  enough  of  the 
for  Expanding  Gases  and  Va-  substance  to  be  discharged  so  that  the 

constant  pressure  of  P2  is  maintained. 

The  quantity  of  energy  and  the  mass  passing  into  the  section  A 
must  be  accounted  for  at  section  B  and  the  relation  of  these 
quantities  will  determine  the  change  of  the  velocity  of  the  sub- 
stance. 

After  uniform  conditions  have  been  established  in  the  nozzle, 
the  same  mass  entering  at  A  must  be  discharged  at  B  during 
the  same  time.  Thus  any  mass  may  be  considered  as  a  work- 
ing basis,  but  as  a  rule  one  pound  of  the  substance  is  used.  All 
formulas  refer  to  one  pound,  unless  another  mass  is  definitely 
stated. 

The  same  quantity  of  energy  discharged  at  B  must  enter  at 
A  unless  heat  is  added  to  or  taken  from  the  substance  between 
the  sections  A  and  B.  Thus  the  general  formula  is  derived: 

Energy  carried  by  substance  at  B  =  energy  carried  by  sub- 
stance at  A  +  energy  added  between  the  sections  A  and  B. 

The  energy  carried  by  the  substance  at  the  entering  or  dis- 
charge end  of  the  nozzle  is  made  up  of  three  quantities:  (i)  the 

150 


FLOW  THROUGH  A  NOZZLE  OR  ORIFICE  151 

amount  of  work  necessary  to  maintain  constant  pressure  at  each 
end  of  the  nozzle;    (2)  the  internal  energy  of  the  substance, 
(3)  the  kinetic  energy  stored  in  the  substance  because  of  the 
velocity  which  it  has  when  passing  the  section. 
The  amount  of  work  necessary  to  maintain  a  constant  pres- 

sure (pounds  per  square  foot)  of  PI  at  A  or  of  P2  at  B  is  —  ^  or 

778 

—  —  ,  where  i\  and  %  are  the  volumes  of  one  pound  of  the  sub- 
778 

stance  at  A  and  B  respectively. 

The  internal  or  "  intrinsic  "  energy  in  B.t.u.  per  pound  (In) 
of  the  substance  passing  A  or  B,  calculating  from  32°  F.  is  for: 

..          .    .,  R  X  (T  -  492)*  f    ,-, 

Air  or  similar  gases,  -  >  (206) 

778X0.40 

Liquid,  h,  (207) 

Liquid  and  Vapor,    h  +  x  I  L  -        -  -  j>  (208) 

/  72       \ 

Superheated  Vapor,  h  +  {L  -  _—  _^J 

0,  (209) 


where  -   -  -  is  the  kinetic  energy  in  B.t.u.  per  pound  of  the 

substance  as  it  passes  a  section,  V  —  velocity  in  feet  per  second 
and  g  =  32.2  feet. 

If  Q  represents  in  B.t.u.  per  pound  the  heat  units  added  to  the 
substance  between  the  sections  A  and  B  (Fig.  45),  the  energy 
equation  for  air  and  similar  gases  can  be  found  by  equating  the 
total  heat  energy  put  in  at  A  plus  the  energy  added  between 
A  and  B  to  the  energy  discharged  at  B.  This  general  formula 
reduces  to 

F2*  -  Fi2  =  2  gf^i  (PlVl  -  P*i)  +  778  Q\ 
Lo-4  J 

or  =  2  g  X  778  [C9  (T,  -  T2)  +  Q].         (210) 

*  See  equation  (61). 


152  FLOW  OF  FLUIDS 

This  thermodynamic  equation  is  the  usual  form  for  the  flow 
of  air  or  similar  gases. 

The  energy  equation  for  superheated  steam  can  be  derived 
by  the  use  of  the  same  general  formula  stated  above,  on  the 
assumption  that  the  substance  is  superheated  at  A  and  B.  This 
formula  reduces  to 


-  [fc  +  L2  +  cp  (rsup.  -  rsat.)2]  +  e,  (211) 

or  F22  -  Fi2  =  2  g  X  778  [Tabular  heat  content,*  -  Tabu- 

lar heat  content?.  -\-  Q].  (212) 

From  a  thermodynamic  standpoint,  the  relation  between  the 
initial  and  final  condition  is  that  of  adiabatic  expansion  when  all 
the  heat  which  disappears  as  such  is  used  in  changing  the  veloc- 
ity, provided  the  nozzle  is  properly  shaped  and  Q  is  zero.  The 
diagram  in  Fig.  46  represents  this  condition  of  affairs  on  a  tem- 
perature-entropy diagram  for  air  and  similar  substances.  The 
area  acdf  is  Cp  (T%  —  492),  and  the  area  abef  is  Cp  (Ti  —  492). 
The  quantity  of  heat  energy  changed  into  kinetic  energy  is  there- 
fore the  area  bcde  and  is  the  difference  between  the  internal 
energy  in  the  substance  at  the  beginning  and  end  of  the  opera- 
tion, together  with  the  excess  of  work  done  to  maintain  the  pres- 
sure of  PI  at  A  over  the  pressure  of  P2  at  B.  The  line  cd  would 
incline  to  the  right  if  heat  were  added  in  the  nozzle,  since  the 
effect  would  be  to  increase  the  velocity  or  increase  the  area  cdeb. 
The  line  cd  would  incline  to  the  left  from  c  if  heat  were  lost, 
since  the  area  cdeb  would  decrease. 

In  Fig.  47  the  diagram  represents  the  conditions  for  super- 
heated vapor.  The  area  aa'cdf  represents  the  heat  required  to 
raise  the  substance  from  a  liquid  at  32°  F.  to  superheated  vapor 
at  the  temperature  of  Jisup.  The  area  ab'bdf  is  the  heat  con- 
tent at  b,  the  final  condition.  The  area  a'cbb'  represents,  there- 
fore, the  heat  available  for  increasing  the  velocity.  The  areas 

*  Tabular  heat  content  means  the  total  heat  of  superheated  steam  as  read 
from  tables  of  the  properties  of  superheated  steam. 


FLOW  THROUGH  A  NOZZLE  OR  ORIFICE 


153 


representing  the  heat  available  for  increasing  the  velocity  in  Fig. 

46  and  Fig.  47  are  shown  by  the  cross-hatched  area  in  Fig.  48 

and  are  really  the  representation  of  the  work  done  (theoretically) 

in  an  engine  giving  such  an  indicator  diagram. 

Evidently  the  greater  the  drop 
in  pressure,  the  greater  will  be  the 
cross-hatched  area  in  Fig.  48  and 
the  greater  will  be  the  velocity, 


T.-  Ent.-  Diagram 
I 


/ 


Tjsat.     / 


FIG.  46.  —  Temperature- 
entropy  Diagram  of  Heat 
Available  in  Air. 


FIG.  47.  —  Temperature-entropy  Dia- 
gram of  the  Heat  Available  in  Superheated 
Steam  for  Increasing  Velocity. 


regardless  of  the  substance.     The  line  ab  in  Fig.  49  represents 

p 
the  velocity  curve  with  V  as  ordinates  and  —  as  abscissas,  but 

since  with  any  substance  ex- 
panding the  weight  of  a  cubic 
foot  decreases  as  the  pressure 
drops,  the  line  cd  will  repre- 
sent to  some  scale  not  here 
determined  the  weight  of  a 
cubic  foot  at  any  discharge 
pressure. 


j.  48.  —  Heat  (Work)  Available  for 
Increasing  Velocity. 


Since  the  product  of  the  area  through  which  the  discharge 
takes  place,  the  velocity  and  the  weight  per  cubic  foot  of  the 


154 


FLOW  OF  FLUIDS 


substance  is  equivalent  to  the  weight  in  pounds  of  the  substance 
discharged  per  second,  the  product  of  the  ordinates  at  any  point 
of  the  curves  ab  and  cd  is  proportional  to  the  weight  discharged 
from  a  pressure  of  PI  to  a  condition  where  the  pressure  is  P2. 
The  line  ceb  represents  this  product.  Evidently  there  is  some 
low  pressure  into  which  the  weight  discharged  per  square  foot 
will  be  a  maximum,  and  this  will  be  the  pressure  corresponding 
to  the  high  point  e  on  the  curve. 


0       .1       .2       .3       .4        .5       .6       .7       .8        .9       1.0 
FIG.  49.  —  Illustrative  Curves  of  Weight,  Discharge  and  Velocity. 

Weight  per  Cubic  Foot.     From  the  formula  for  adiabatic  ex- 
pansion the  weight  per  cubic  foot  can  be  obtained  if  the  sub- 
stance is  similar  to  air.*     The  general  formula  (applied  to  air)  is 
p^i-40  =  Ptfz1-™,  (213) 

which  can  be  reduced  to 


P21'4  X  Pi,  i 


0.286 


>  or     —  = 


RTl 


(214) 


Pi"  X  RTi 

*  The  exponent  in  the  formula  is  the  ratio  of  the  specific  heats  (of  air  in  this 


case). 


FLOW  OF  AIR  THROUGH  AN  ORIFICE  155 

which  is  the  weight  in  pounds  per  cubic  foot  of  discharge.  Pres- 
sures are  in  pounds  per  square  foot.  If  the  supply  to  the  nozzle 
is  from  a  large  reservoir  so  that  Vi  can  be  taken  as  zero  then 
the  discharge  velocity  is 


0.4 


F,=  109.6 

All  quantities  on  the  right-hand  side  of  this  equation  must  be 
obtained  from  the  data  of  tests.  Weight  in  pounds  discharged 
through  the  area  A  (in  square  feet)  is 


0-286 


Maximum   Discharge.      This   weight   is   a   maximum   when 

dW 

-  =  o  or  when  P2  =  0.525  PI.     The  maximum  quantity  of  air 


will  be  discharged  when  the  low  pressure  is  52.5  per  cent  of  the 
high  pressure. 

Shape  of  Nozzle.     See  page  167,  on  the  flow  of  steam. 

Flow  of  Air  through  an  Orifice.  Air  under  comparatively  high 
pressure  is  usually  measured  in  practice  by  means  of  pressure 
and  temperature  observations  made  on  the  two  sides  of  a  sharp- 
edged  orifice  in  a  diaphragm.  The  method  requires  the  use  of 
two  pressure  gages  on  opposite  sides  of  the  orifice  and  a  ther- 
mometer for  obtaining  the  temperature  ti  at  the  initial  or  higher 
pressure  pi.  The  flow  of  air  w,  in  pounds  per  second,  may  then 
be  calculated  by  Fliegner's  formulas  : 

w  =  0.530  X  f  X  a  -Jjp=  when  pi  is  greater  than  2  p2,        (217) 

w  =  i.  060  X  f  X  a  1/P2  (Pi  ~  P?)  when  pi  is  less  than  2  &,  (218) 
V 


156  FLOW  OF  FLUIDS 

where  a  is  the  area  of  the  orifice  in  square  inches,  f  is  a  coeffi- 
cient, TI  is  the  absolute  initial  temperature  in  degrees  Fahren- 
heit at  the  absolute  pressure  pi  in  the  "  reservoir  or  high-pressure 
side"  and  p2  is  the  absolute  discharge  pressure,  both  in  pounds 
per  square  inch.  When  the  discharge  from  the  orifice  is  directly 
into  the  atmosphere,  p2  is  obviously  the  barometric  pressure. 

Westcott's  and  Weisbach's  experiments  show  that  the  values 
of  f  are  about  0.925  for  equation  (217)  and  about  0.63  for  equa- 
tion (218). 

For  small  pressures  it  is  often  desirable  to  substitute  manom- 
eters for  pressure  gages.  One  leg  of  a  U-tube  manometer  can  be 
connected  to  the  high-pressure  side  of  the  orifice  and  the  other  leg 
to  the  low-pressure  side.  Valves  or  cocks  are  sometimes  inserted 
between  the  manometer  and  the  pipe  in  which  the  pressure  is  to 
be  observed  for  the  purpose  of  " dampening"  oscillations.  This 
practice  is  not  to  be  recommended  as  there  is  always  the  possi- 
bility that  the  pressure  is  being  throttled.*  A  better  method  is 
to  use  a  U-tube  made  with  a  restricted  area  at  the  bend  between 
the  two  legs.  This  will  reduce  oscillations  and  not  affect  the 
accuracy  of  the  observations. 

Discharge  from  compressors  and  the  air  supply  for  gas  engines 
are  frequently  obtained  by  orifice  methods. 

When  pi  —  p2  is  small  compared  with  pi,  the  simple  law  of 
discharge  f  of  fluids  can  be  used  as  follows: 

*  Report  of  Power  Test  Committee,  Journal  A. S.M .E.,  Nov.,  1912,  page"i6g5. 
f  If  the  density  is  fairly  constant, 

144pi       vi«       144  p2   ,   v0a 
~T""i"2g"       s       "t"2g> 

where  Vi  is  the  velocity  in  feet  per  second  in  the  "approach"  to  the  orifice,  and  v0  is 
the  velocity  in  the  orifice  itself.  Since  Vi  should  be  very  small  compared  with  v0, 

V02  _  144  (Pl  -  p2) 
2g"  s 


Vo  =  4/2  g  X  144  (p:  -  p2). 


FLOW  OF  AIR  THROUGH  AN  ORIFICE  157 


fa 


a    /  - 

w  =  -  -V2  g  X  144  (Pi  -  P2)  s,  (219) 

144 

where  f  is  a  coefficient  from  experiments,  g  is  the  acceleration 
due  to  gravity  (32.2),  and  s  is  the  unit  weight  of  the  gas  meas- 
ured, in  pounds  per  cubic  foot,  for  the  average  of  the  initial  and 
final  conditions  of  temperature  and  pressure.  If  the  difference 
in  pressure  is  measured  in  inches  of  water  h  with  a  manometer, 
then 

144  (pi  —  p2)  =  -    -  X  h  (expressed  in  terms^of  pounds  per  square 

foot), 


w  =  --  y  2  ghs  X  -    -  (pounds  per  second),     (220) 
144  12 

where  62.4  is  the  weight  of  a  cubic  foot  of  water  (density)  at 
usual  "room"  temperatures. 

This  equation  can  also  be  transformed  so  that  a  table  of  the 
weight  of  air  is  not  needed,  since  by  elementary  thermodynamics 
144  pv  =  53.3  T,  where  v  is  the  volume  in  cubic  feet  of  one 
pound  and  T  is  the  absolute  temperature  in  Fahrenheit.  Since 
v  is  the  reciprocal  of  s,  then 

s  =  144  p  ^  53.3  T, 


and  w  =  0.209  fa  y-  (221) 

TI 

Here  p  and  T  should  be  the  values  obtained  by  averaging  the 
initial  and  final  pressures  and  temperatures.  Great  care  should 
be  exercised  in  obtaining  correct  temperatures.  For  accurate 
work,  corrections  of  s  for  humidity  must  be  made.* 

For  measurements  made  with  orifices  with  a  well-rounded 
entrance  and  a  smooth  bore  so  that  there  is  practically  no  con- 


or  w  =  f  a  V2  g  X  144  (px  -  p2)s. 

Professor  A.  H.  Westcott  has  computed  from  accurate  experiments  that  the 
value  of  the  coefficient  f  in  these  equations  is  approximately  0.60. 

*  Tables  of  the  weight  of  air  are  given  on  page  181  and  tables  of  humidity  on 
page  368  in  Meyer's  Power  Plant  Testing  (2d  Edition). 


158  FLOW   OF   FLUIDS 

traction  of  the  jet  the  coefficient  f  in  equations  (217)  and  (218) 
may  be  taken  as  0.98.  In  the  rounding  portion  of  the  entrance 
to  such  a  nozzle  the  largest  diameter  must  be  at  least  twice  the 
diameter  of  the  smallest  section.  For  circular  orifices  with 
sharp  corners  Professor  Dalby  *  stated  that  the  coefficient  for 
his  sharp-edged  orifices  in  a  thin  plate  of  various  sizes  from  i  inch 
to  5  inches  in  diameter  was  in  all  cases  approximately  0.60; 
and  these  data  agree  very  well  with  those  published  by  Durley.f 

When  p2  -T-  pi  =  0.99  the  values  obtained  with  this  coefficient 
are  in  error  less  than  J  per  cent;  and  when  p2  -f-  pi  =  0.93  the 
error  is  less  than  2  per  cent. 

Receiver  Method  of  Measuring  Air.  None  of  the  preceding 
methods  are  adaptable  for  measuring  the  volume  of  air  at  high 
pressures  as  in  the  case  of  measuring  the  discharge  in  tests  of 
air  compressors.  Pumping  air  into  a  suitably  strong  receiver 
is  a  method  often  used.  The  compressor  is  made  to  pump 
against  any  desired  pressure  which  is  kept  constant  by  a  regu- 
lating valve  on  the  discharge  pipe: 

PI  and  ?2  =  absolute  initial  and  final  pressures  for  the  test, 
pounds  per  square  inch. 

TI  and  T2  =  mean  absolute  initial  and  final  temperatures,  de- 
grees Fahrenheit. 

Wi  and  W2  =  initial  and  final  weight  of  air  in  the  receiver, 
pounds. 

V  =  volume  of  receiver,  cubic  feet. 

pty  =  wRTi,  and  P2V  =  w2RT2,  where  R  is  the  constant 
53.3  for  air,  then  weight  of  air  pumped 


53-3 

In  accurate  laboratory  tests  the  humidity  of  the  air  entering 
the  compressor  should  be  measured  in  order  to  reduce  this 

*  Engineering  (London),  Sept.  9,  1910,  page  380,  and  Ashcroft  in  Proc.  Insti- 
tution of  Civil  Engineers,  vol.  173,  page  289. 

f  Transactions  American  Society  of  Mechanical  Engineers,  vol.  27  (1905), 
page  193. 


FLOW  OF  VAPORS  159 

weight  of  air  to  the  corresponding  equivalent  volume  at  atmos- 
pheric pressure  and  temperature. 

The  principal  error  in  this  method  is  due  to  difficulty  in  measur- 
ing the  average  temperature  in  the  receiver.  Whenever  practi- 
cable the  final  pressure  should  be  maintained  in  the  receiver  at 
the  end  of  the  test  until  the  final  temperature  is  fairly  constant. 

Flow  of  Vapors.  In  Fig.  45  suppose  the  sections  A  and  B 
are  so  proportioned  that  the  velocity  of  the  substance  passing 
section  A  is  the  same  as  that  at  section  B.  Such  a  condition 
might  arise  in  a  calorimeter  or  in  the  expansion  of  ammonia 
through  a  throttling  or  expansion 


T-Ent,-Diagram 


P 

valve,  as  in  an  ice  machine.  The 
pressure  at  A  will  be  PI  which  is 
greater  than  the  pressure  at  B  of  P2. 
Fig.  50  represents  the  entropy  dia- 
gram for  such  a  condition.  As  the 
pressure  falls  from  PI  to  P2  the  maxi- 
mum heat  available  to  produce  veloc- 
ity through  the  nozzle  is  the  area 
acde.  The  value  of  the  quality,  rep- 
resented by  the  symbol  x  for  the 

substance   after    leaving    the    nozzle  FlG'  so.  -Diagram  for  no  Ve- 

locity Change. 
corresponds  to  that  of  point  c   and 

the  area  acde  is  the  excess  kinetic  energy  represented  by  the 
increased  velocity.  This  excess  kinetic  energy  is  destroyed  by 
coming  into  contact  with  the  more  slowly  moving  particles  at 
B  and  with  the  sides  of  the  vessel.  The  area  acde  is  equal  to 
(ha  +  xaLa)  —  (hc  -f  xcLc)  and  the  relation  of  xa  to  xc  is  obvi- 
ously adiabatic.  The  area  ohdbf  equals  area  oheag  (thus  the 
heat  content  at  b  is  the  same  as  at  a).  The  location  of  b  can  be 
found  as  follows: 

(223) 


The  curve  shown  in  Fig.  51  represents  the  discharge  of  a  mix- 
ture of  steam  and  water  (x  =  0.6  at  100  pounds  per  square  inch 
absolute  pressure)  into  a  vessel  having  the  pressures  shown.  The 


i6o 


FLOW   OF   FLUIDS 


points  on  this  curve  cannot  be  determined  by  entropy  tables. 
At  100  pounds  per  square  inch  pressure  the  total  heat  of  the  wet 
steam  is  h  +  0.6  L  or  298.5  +  0.6  X  887.6  =  831.1  B.t.u.  per 
pound. 


1.75 


1.50 


t  smallest  s 
li>s.  per  sec. 

i 


pe 


§ 


0     10     20     30    40     50     60     70    80     90    100 
Discharge  Pressure Jn.  Ibs.  per  sq.  in. 

FIG.  51.  —  Discharge  of  Steam  Under  Various  Pressures. 

At  60  pounds  per  square  inch  absolute  pressure  the  total  heat 
may  be  found  by  the  use  of  the  entropy  diagram  shown  in 
Fig.  52.     The  entropy  values  are  taken  directly  from  steam 
tables.     The  entropy  for  the  initial  point  is,  then, 
ab  =  0.4748  +  0.6  X  1.1273  =  1.1512. 

The  distance 

dc  =  ab  -  0.4279  =  0.7233, 

and  x  Sit  60  pounds  is: 

dc       _  0.7233 


=  °-595- 
1.2154      1.2154 

The  total  heat  at  60  pounds  pressure  is 

h  +  0.595  L  =  262.4  +  0.595  X  914.3 


805.4. 


FRICTION  LOSS  IN  A  NOZZLE 


161 


The  velocity  of  flow  is 


V2  x  32.2  x  778  (831.1  -  805.4)  =  1135  feet  per  second. 
The  volume  of  one  pound  is 

0.016  (i.o  —  0.595)  +  7.166  X  0.595  =  4.27  cubic  feet, 
and  the  weight  per  cubic  foot  is 

--  =  0.2342  pound. 
4.27 

The  weight  discharged  per  square 
inch  per  second  is 

1135  X  0.2342 

-  =  1.85  pounds. 
144 

Velocity  of  Flow  as   Affected   by 
Radiation.     Fig.  53  shows  the  radia- 

°  FIG.    52.  •  —  Tempera  ture-en- 

tion  losses.     The  condition  at  entrance  tropy  Diagram  for  Calculating 
is  represented  at  a  and  the  area  acde  the   Weight   of   Discharge  of 
represents  the  quantity  of  heat  lost  Steam- 
by  radiation.    Area  aefg  represents  the  velocity  change  while 
the  point  e  represents  the  condition  of  the  moving  substance. 

If,  after  passing  through  the  nozzle, 
the  velocity  is  reduced  to  that  of  en- 
trance, a  point  located  as  at  b  will 
represent  the  condition  of  the  sub- 
stance. This  point  would  be  so  located 
that 

area  aefg 
eb  =  -  —  — 


T.-Ent.-  Diagram 


d\    c\ 


(224) 


Friction  Loss  in  a  Nozzle.    Fig.  54 

shows  the  friction  loss.  The  energy 
converted  into  heat  by  friction  varies 
with  the  square  of  the  velocity.  In 

FIG.  53.  — Diagram  illustrating  this  figure,  a  is  the  initial  condition 

and  aefg  is  the  energy  available  for 

change  in  velocity  provided  there  is  no  frictional  loss.    The  ratio 

of  the  areas  acde  to  aefg  is  the  proportional  loss  by  friction. 

The  point  c  represents  the  condition  of  the  substance  at  the 


l62 


FLOW  OF  FLUIDS 


return  of   the  friction  heat  to  the  substance.      The  heat  is 
returned  in  exactly  the  same  way  as  if  it  came  from  an  outside 

source.  The  distance  ch  is  the  area 
acde  -=-  Lc.  The  area  edfg  represents 
the  energy  expended  in  the  velocity 
change  and  the  point  h  represents  the 
state  of  the  substance  at  the  point  of 
discharge. 

This  condition  is  found  existing  in 
the  fixed  nozzle  of  most  turbines. 
Point  a  represents  the  condition  on 
the  high-pressure  side  of  the  nozzle 
and  point  h,  the  low-pressure  side. 
The  absolute  velocity  of  discharge  is 


T.-  Ent.-  Diagram 


M 


n  I    d 


FIG.  54.  —  Diagram  Illustrating  really  caused  by  the  energy  repre- 
Friction  Loss  in  Nozzle.        sented  by  the  area  edfg. 

Impulse  Nozzles.  Suppose  that  the  substance  is  discharged 
with  an  absolute  velocity  corresponding  to  the  area  edfg  (Fig. 
54),  and  that  it  passes  into  a  mov- 
ing nozzle,  having  the  same  pressure 
on  the  discharge  as  on  the  intake 
side.  The  energy  represented  by 
the  area  edfg  would  be  used  up  in 
the  following  ways:  (i)  by  the  fric- 
tion hi  moving  nozzle;  (2)  residual 
absolute  velocity;  (3)  and  energy 
used  in  driving  the  moving  nozzle 
against  the  resistance. 

Fig.  55  shows  the  quantities  used 
up  by  each.  Point  a  represents  the 
condition  of  the  substance  before 
passing  into  the  fixed  nozzle  while 
point  h  shows  its  condition  leaving 
the  fixed  nozzle,  the  velocity  corre- 
sponding to  the  area  edfg.  Area  klde  represents 


T.-  Ent.-  Diagram 


used  up  in  friction  in  the  moving  nozzle;    area 


the  energy 
klnm  residual 


TURBINE  LOSSES  163 

velocity  after  leaving  moving  nozzle  and  area  mnfg  represents 
useful  work  used  in  moving  the  nozzle  against  its  resistance. 
The  condition  of  the  substance  leaving  the  nozzle  is  shown  at 
q  and  not  at  h,  the  distance  h  —  q  being  the  area  edlk  divided 
by  L*.  The  substance  leaves  the  moving  nozzle  with  a  velocity 
corresponding  to  the  area  klnm  and  it  will  have  done  work 
corresponding  to  the  area  mnfg. 

Turbine  Losses.  Fig.  56  is  a  simple  velocity  diagram  show- 
ing, for  an  impulse  nozzle  such  as  occurs  in  many  turbines,  the 
relative  value  of  those  various  losses.  A  is  a  stationary  nozzle 


*— 98) 


€V 

FIG.  56.  —  Impulse  Nozzle  and  Velocity  Diagrams. 

discharging  against  the  movable  blades  B.  The  path  of  the  steam 
is  shown  by  the  dotted  line.  The  line  db  marked  v  represents 
the  velocity  of  discharge  of  the  stationary  nozzle,  which  makes 
an  Bangle  a  with  the  direction  of  motion  of  the  moving  blades. 
Call  &  the  velocity  of  the  moving  blades,  then  h  is  the  amount 
and  direction  of  the  relative  velocity  of  the  steam  over  the 
surface  of  the  moving  blades.  It  loses  a  portion  of  this  velocity 
as  it  passes  over  the  surface  of  the  blades  and  ///  becomes  the 
actual  relative  velocity  of  discharge.  The  direction  of  Ik  is  de- 
•  termined  by  the  discharge  edge  of  the  moving  blades,  the  angles 
a  and  0  being  as  shown.  The  residual  absolute  velocity  is  rep- 
resented by  r. 
The  total  energy  equivalent  of  the  velocity  developed 

in  B.t.u.  per  pound  =  - —  (225) 

2  8 


164 


FLOW   OF  FLUIDS 


The  residual  energy  per  pound      =  - 


(226) 


in 


o\  e 

I 


d 


Reaction  Nozzles.  When  the  substance  leaving  the  station- 
ary nozzle  passes  into  a  moving  nozzle  having  the  pressure  at 
the  intake  greater  than  at  the  discharge,  the  conditions  differ 
from  those  just  discussed.  The  velocity  in  this  case  is  changed 
in  passing  through  the  moving  nozzle.  In  the  equations  given 
in  the  previous  discussion  it  was  assumed  that  the  moving  nozzles 

were  entirely  rilled  with  the  sub- 
stance, and  when  partly  filled  in  the 
expanding  portion,  coefficients  of 
correction  were  applied,  but  in  this 
case  the  nozzles  should  be  so  de- 
signed that  the  substance  entirely 
fills  them,  as  the  corrections  are  un- 
known. 

In  Fig.  57  the  lines  of  Fig.  55  are 
reproduced  together  with  those  re- 
lating directly  to  the  reaction  nozzle. 
Point  a  corresponds  to  the  condition 
on  entering  the  stationary  nozzle, 


T.-Ent.- Diagram 


FiG.57.-DiagramofHeatLosses 
in  a  Steam  Nozzle  (Reaction). 


point  h  the  condition  on  leaving  it 
with  a  velocity  corresponding  to  the 
area  edfg.  Point  k  represents  the  pressure  at  the  discharge  end 
of  the  moving  nozzle,  and  if  no  friction  losses  or  impact  loss  occur 
in  the  moving  nozzle,  point  k  would  represent  the  condition  of  the 
discharged  substance  and  the  area  egmkhd  would  be  accounted 
for  as  useful  work  done  and  residual  velocity.  But  since  friction 
losses  and  impact  loss  do  occur  a  portion  of  this  area  edhkno  can 
be  set  aside  to  represent  these  losses,  a  portion  noqp  represents 
the  residual  velocity,  while  the  remaining  area  pqgm  represents 
the  useful  work  done. 
The  condition  of  the  substance  leaving  the  moving  nozzle  is 

,      ,.  ,  ,.      area  edhkno       . 

given  by  k,  the  distance  kl  being Area  onpq  rep- 
resents the  residual  velocity. 


INJECTORS  165 

Coefficient  of  Flow.  Few  experiments  have  been  carried  on  for 
determining  the  flow  of  steam  in  nozzles  proportioned  for  maxi- 
mum discharge.  For  a  nozzle  having  a  well  rounded  entrance 
and  with  the  parallel  portion  of  least  diameter  from  0.25  to  1.5 
times  the  length  of  the  converging  entrance,  the  coefficient  of 
discharge  is  about  1.05.  For  properly  shaped  entrances  and  for 
areas  of  orifices  between  0.125  square  inch  and  0.75  square  inch 
the  coefficient  of  discharge  varies  from  0.94,  the  two  pressures 
being  nearly  alike,  to  unity,  the  ratio  of  the  pressures  being  0.57. 
For  an  orifice  through  a  thin  plate  the  coefficient  is  about  0.82, 
the  ratio  of  the  pressures  being  0.57. 

Injectors.     In  an  injector,  steam  enters  at  A  in  Fig.  58  at 
the  pressure  of  the  supply.     The  quantity  of  water  entering  at 
C,  the  cross-section  of  the  pipe,  and  the  pressure  of  the  water 
determine  the  pressure  at  B.    At  D 
the  pressure  should  be  zero  (atmos-        ^5?         "^  o 
pheric)  or  equal  to  the  pressure  in 
the  water  supply  pipe  to   which  D    FIG.  58.  -  Essential  Parts  of  an 
may  be  connected.     The  total  hy- 
draulic head  should  exist  as  velocity  head  at  this  point.    At  E 
the  pressure  should  be  sufficient  to  raise  the  check  valve  into 
the  boiler  and   the  velocity  sufficient  to   carry  the   intended 
supply  into  the  discharge  pipe. 

The  shape  of  the  nozzle  from  A  to  B  should  be  such  as  to  con- 
vert the  erfergy  in  the  steam  at  A  into  velocity  at  B.  At  B  the 
water  and  steam  meet,  condensing  the  steam,  heating  the  water 
and  giving  to  the  water  a  velocity  sufficient  to  carry  it  through 
the  nozzle  B-D. 

All  the  energy  accounted  for  at  A  and  C  must  be  accounted  for 
at  E.  The  heat  lost  by  radiation  may  be  neglected.  The  veloc- 
ity at  any  section  of  the  nozzle  equals 

volume  passing  in  cubic  feet 

area  of  section  in  square  feet ' 

aV 
or  VA  =  -— >  where  a  =  area  at  any  section  corresponding  to 

velocity  V  and  a  A  =  area  at  section  A. 


l66  FLOW  OF  FLUIDS 

Weight  of  Feed  Water  Supplied  by  an  Injector  per  Pound  of 
Steam.  Assuming  the  steam  supply  to  be  dry  the  heat  units 
contained  in  the  steam  and  feed  water  per  pound  and  the  heat 
in  the  mixture  of  steam  and  feed  water  per  pound  may  be  easily 
calculated. 

Knowing  the  rise  of  temperature  of  the  water  passing  through 
the  injector  and  neglecting  radiation  losses,  the  pounds  of  feed 
water  supplied  per  pound  of  steam  used  by  the  injector  may  be 
obtained.  Thus: 

Heat  units  lost  by  steam  =  Kinetic  energy  of  jet  +  Heat 
units  gained  by  feed  water. 

The  kinetic  energy  of  jet  may  be  neglected  since  it  is  very 
small,  then, 

H  -  h,  =  w(hm-hf) 

H  —  ht  /      \ 

w  =  h^J/         ;  .       (227) 

where  w  =  the  weight  of  feed  water  lifted  per  pound  of  steam. 
km  =  heat  of  liquid  of  mixture  of  condensed  steam  and 

feed  water. 
hf   =  heat  of  liquid  of  entering  feed  water. 

Thermal  Efficiency  of  Injector.  The  thermal  efficiency  of  an 
injector  neglecting  radiation  losses  is  unity.  All  the  heat  ex- 
pended is  restored  either  as  work  done  or  in  heat  returned  to  the 
boiler. 

Mechanical  Efficiency  of  Injector.  The  mechanical  work  per- 
formed by  the  injector  consists  in  lifting  the  weight  of  feed  water 
and  delivering  it  into  the  boiler  against  the  internal  pressure. 
The  efficiency,  considering  the  injector  as  a  pump,  is 

Work  done 

B.t.u.  given  up  by  steam  to  perform  the  work 
or 


where  U  =  [wls  +  (w  +  i)  lp]  -f-  778  (in  heat  units). 


FLOW  OF  STEAM  THROUGH  NOZZLES 


167 


IP  =  pressure  head  corresponding  to  boiler  gage  pressure, 

in  feet. 

ls  =  suction  head  in  feet. 
w  =  pounds  of  water  delivered  per  pound  of  steam. 

Orifice  Measurements  of  the  flow  of  steam  are  sometimes  used 
for  ascertaining  the  steam  consumption  of  the  " auxiliaries"  in  a 
power  plant.  This  method  commends  itself  particularly  because 
of  its  simplicity  and  accuracy.  It  is  best  applied  by  inserting  a 
plate  J  inch  thick  with  an  orifice  one  inch  in  diameter,  with  square 
edges,  at  its  center,  between  the  two  halves  of  a  pair  of  flanges 
on  the  pipe  through  which  the  steam  passes.  Accurately  cali- 
brated steam  gages  are  required  on  each  side  of  the  orifice  to 
determine  the  loss  of  pressure.  The  weight  of  steam  for  the 
various  differences  of  pressure  may  be  determined  by  arranging 
the  apparatus  so  that  the  steam  passing  through  the  orifice  will 
be  discharged  into  a  tank  of  water  placed  on  a  platform  scales. 
The  flow  through  this  orifice  in  pounds  of  dry  saturated  steam 
per  hour  when  the  discharge  pressure  at  the  orifice  is  100  pounds 
by  the  gage  is  given  by  the  following  table: 


Pressure  drop, 
Ibs.  per  sq.  in. 

Flow  of  dry  steam 
per  hour,  Ibs. 

Pressure  drop, 
Ibs.  per  sq.  in. 

Flow  of  dry  steam 
per  hour,  Ibs. 

i 

430 

5 

1560 

i 

6l5 

IO 

2180 

2 

93° 

15 

2640 

3 

I2OO 

20 

3050 

I4OO 

. 

Flow  of  Steam  through  Nozzles.  The  weight  of  steam  dis- 
charged through  any  well-designed  nozzle  with  a  rounded  inlet, 
similar  to  those  illustrated  in  Figs.  59  and  60,  depends  only  on 
the  initial  absolute  pressure  (Pi),  if  the  pressure  against  which 
the  nozzle  discharges  (P2)  does  not  exceed  0.58  of  the  initial 
pressure.  This  important  statement  is  well  illustrated  by  the 
following  example.  If  steam  at  an  initial  pressure  (Pi)  of  100 
pounds  per  square  inch  absolute  is  discharged  from  a  nozzle,  the 


i68 


FLOW  OF  FLUIDS 


weight  of  steam  flowing  in  a  given  time  is  practically  the  same 
for  all  values  of  the  pressure  against  which  the  steam  is  dis- 
charged (P2),  which  are  equal  to  or  less  than  58  pounds  per  square 
inch  absolute. 

If,  however,  the  final  pressure  is 
more  than  0.58  of  the  initial,  the  weight 
of  steam  discharged  will  be  less,  nearly 
in  proportion  as  the  difference  between 
the  initial  and  final  pressures  is  re- 
duced. 

The  most  satisfactory  and  accurate 
formula  for  the  "  constant  flow  "  con- 
dition, meaning  when  the  final  pressure 
is  0.58  of  the  initial  pressure  or  less,  is 

the  following.  due  to  Grashof,*  where w 
is  the  flow  of  steam  f  (initially  dry 
saturated)  in  pounds  per  second,  AQ  is  the  area  of  the  smallest 
section  of  the  nozzle  in  square  inches,  and  PI  is  the  initial  abso- 
lute pressure  of  the  steam  in  pounds  per  square  inch, 


or,  in  terms  of  the  area, 


A,  = 


60  w 
Pi*1 ' 


(229) 


(230) 


*  Grashof,  Theoretische  Maschinenlehre,  vol.  i,  iii;   Hiltte  Taschenbuch,  vol.  i, 
page  333.     Grashof  states  the  formula, 

w  =  0.01654  4oPr9696, 

but  the  formula  given  in  equation  (229)  is  accurate  enough  for  all  practical  uses. 

|  Napier's  formula  is  very  commonly  used  and  is  accurate  enough  for  most 
calculations.    It  is  usually  stated  in  the  form 

A<J>i 

w  = , 

70 

where  w,  PI,  and  AQ  have  the  same  significance  as  in  Grashof  s  formula.    The 
following  formula  is  given  by  Rateau,  but  is  too  complicated  for  convenient  use: 

w  =  0.001  AoPi  [15-26  —  0.96  (log  Pi  +  log  0.0703)]. 
Common  or  base  10  logarithms  are  to  be  used  in  this  formula. 


FLOW  OF  STEAM  THROUGH  NOZZLES 


I69 


These  formulas  are  for  the  flow  of  steam  initially  dry  and 
saturated.  An  illustration  of  their  applications  is  given  by  the 
following  practical  example. 

Example.  The  area  of  the  smallest  section  (^40)  of  a  suitably 
designed  nozzle  is  0.54  square  inch.  What  is  the  weight  of  the 


rffh 


DeLaval  Type. 


Curtis  Type. 

FIG.  60.  —  Examples  of  Standard  Designs  of  Nozzles. 

flow  (w)  of  dry  saturated  steam  per  second  from  this  nozzle  when 
the  initial  pressure  (Pi)  is  135  pounds  per  square  inch  absolute 
and  the  discharge  pressure  (P2)  is  15  pounds  per  square  inch  ab- 
solute? 
Here  P2  is  less  than  0.58  Pi  and  Grashof's  formula  is  applicable, 


170  FLOW  OF  FLUIDS 

oX 
or,         w  =     ° 

, 

0.54  x  116.5  * 

w  =     J^     ^—  =  1.049  pounds  per  second. 

oo 

When  steam  passes  through  a  series  of  nozzles  one  after  the 
other  as  is  the  case  in  many  types  of  turbines,  the  pressure 
is  reduced  and  the  steam  is  condensed  in  each  nozzle  so  that 
it  becomes  more  moist  each  time.  In  the  low-pressure  nozzles 
of  a  turbine,  therefore,  the  steam  may  be  very  wet  although 
initially  it  was  dry.  Turbines  are  also  sometimes  designed  to 
operate  with  steam  which  is  initially  wet,  and  this  is  usually  the 
case  when  low-pressure  steam  turbines  are  operated  with  the 
exhaust  from  non-condensing  reciprocating  engines.  In  all  these 
cases  the  nozzle  area  must  be  corrected  for  the  moisture  in  the 
steam.  For  a  given  nozzle  the  weight  discharged  is  greater 
for  wet  than  for  dry  steam;  but  the  percentage  increase  in  the 
discharge  is  not  nearly  in  proportion  to  the  percentage  of  mois- 
ture as  is  often  stated.  The  general  equation  for  the  theoretic 
discharge  (w}  from  a  nozzle  is  in  the  form  | 

*  The  flow  (w)  calculated  by  Napier's  formula  for  this  example  is  w  =  — — — 

70 
=  1.041  pounds  per  second. 

t  The  general  equation  for  the  theoretic  flow  is 


W  =  AQ 


2  *+i-i 

H\I  „(?•}'  I 
(p      -I 


where  the  symbols  w,  A0,  PI,  and  g  are  used  as  in  equations  (228)  and  (229).  P2 
is  the  pressure  at  any  section  of  the  nozzle,  vi  is  the  volume  of  a  pound  of  steam  at 
the  pressure  PI,  and  k  is  a  constant.  The  flow,  w,  has  its  maximum  value  when 

2  t+i 

•    .,        (Sf-GY 

is  a  maximum.    Differentiating  and  equating  the  first  differential  to  zero  gives 

i 

PJ.    (   2   y-i 

PI    U  +  i/ 

P2  is  now  the  pressure  at  the  smallest  section,  and  writing  for  clearness  P0  for 
PZ,  and  substituting  this  last  equation  in  the  formula  for  flow  (w)  above,  we  have 


FLOW  OF   STEAM  171 

,  (23I) 


where  PI  is  the  initial  absolute  pressure  and  Vi  is  the  specific  vol- 
ume (cubic  feet  in  a  pound  of  steam  at  the  pressure  PI).  Now, 
neglecting  the  volume  of  the  water  in  wet  steam,  which  is  a  usual 
approximation,  the  volume  of  a  pound  of  steam  is  proportional 
to  the  quality  (x\).  For  wet  steam  the  equation  above  becomes 
then 

-  (232) 


J 

V 


The  equation  shows,  therefore,  that  the  flow  of  wet  steam  is 
inversely  proportional  to  the  square  root  of  the  quality  (xi). 
Grashof  's  equations  can  be  stated  then  more  generally  as 

'  (333) 

60   W    VXI  f  N 

A0  =  —  p^  —  (234) 

These  equations  become  the  same  as  (229)  and  (230)  for  the 
case  where  x\  =  i. 

Flow  of  Steam  when  the  Final  Pressure  is  more  than  0.58  of 
the  Initial  Pressure.  For  this  case  the  discharge  depends  upon 
the  final  pressure  as  well  as  upon  the  initial.  No  satisfactory 
formula  can  be  given  in  simple  terms,  and  the  flow  is  most  easily 
calculated  with  the  aid  of  the  curve  in  Fig.  61,  due  to  Rateau. 


Now  regardless  of  what  the  final  pressure  may  be,  the  pressure  (P0)  at  the  smallest 
section  of  a  nozzle  (A0)  is  always  nearly  0.58  Pi  for  dry  saturated  steam.  Making 
then  in  the  last  equation  P0  =  0.58  PI  and  putting  for  k  Zeuner's  value  of  1.135 
for  dry  saturated  steam,  we  may  write  in  general  terms  the  form  stated  above, 


where  K  is  another  constant.    See  Zeuner's  Theorie  der  Turbinen,  page  268  (Ed.  of 
1899). 


172 


FLOW  OF  FLUIDS 


This  curve  is  used  by  determining  first  the  ratio  of  the  final  to 
p 

the  initial  pressure  -=£  and  reading  from  the  curve  the  corre- 
"i 

spending  coefficient  showing  the  ratio  of  the  required  discharge 
to  that  calculated  for  the  given  conditions  by  either  of  the  equa- 
tions (229)  or  (233).  The  coefficient  from  the  curve  times  the 


Eatlo  of  Final  to  Initial  Pressure--!?. , 


FIG.  61. —  Coefficients  of  the  Discharge  of  Steam  when  the  Final  Pressure  is 
Greater  than  0.58  of  the  Initial  Pressure. 

flow  calculated  from  equations  (229)  or  (233)  is  the  required  re- 
sult. Obviously  the  discharge  for  this  condition  is  always  less 
than  the  discharge  when  the  final  pressure  is  equal  to  or  less 
than  0.58  of  the  initial. 

Length  for  Nozzles.  The  length  of  the  nozzle  is  usually  made 
to  depend  only  on  the  initial  pressure.  In  other  words,  the 
length  of  a  nozzle  for  150  pounds  per  square  inch  initial  pressure 
is  usually  made  the  same  for  a  given  type  regardless  of  the  final 
pressure.  And  if  it  happens  that  there  is  crowding  for  space, 
one  or  more  of  the  nozzles  may  be  made  a  little  shorter  than 
the  others. 


UNDER-  AND  OVER-EXPANSION 


173 


Designers  of  De  Laval  nozzles  follow  practically  the  same 
"  elastic  "  method.  The  divergence  of  the  walls  of  non-con- 
densing nozzles  is  about  3  degrees  from  the  axis  of  the  nozzle, 
and  condensing  nozzles  for  high  vacuums  may  have  a  divergence 
of  as  much  as  6  degrees  *  for  the  normal  rated  pressures  of  the 
turbine. 

One  of  the  authors  has  used  successfully  the  following  empirical 
formula  to  determine  a  suitable  length,  L,  of  the  nozzle  between 
the  throat  and  the  mouth  (in  inches) : 

(235) 

where  A0  is  the  area  at  the  throat  in  square  inches. 

Under-  and  Over-expansion.  The  best  efficiency  of  a  nozzle  is 
obtained  when  the  expansion  required  is  that  for  which  the  nozzle 


TToZzle  Velocfly  IxwrfDna  tff 
TJnder-or  Over-E*panslou 

O  U>  •*>>  Ci  00 

\ 

/ 

/ 

/ 

/ 

/ 

/ 

/ 

/ 

*x. 

/ 

~^ 

^^. 

z 

Z 

] 

P 

i 

•-i 

ere 

«: 

J 

•lit- 

(Ml 

20       15 
ge  Nozzle  is  to 
h  (Under  Expi 

~  —  » 

1 

o  S 

••i 

JJ 

0        i 

Halt' 

Ml) 

J-n 

5         1 

Perc< 
at 

JLl 

0  1 

aitage 
Mouth  ( 

I 

Voz 
Ovc 

~ 
!•] 

rl 

j 

1  to 
spa. 

1 

oL. 

LUiO 

S 

irgs 

4 

i 

^ 

r~ 

FIG.  62.  —  Curve  of  Nozzle  Velocity  Loss. 

was  designed,  or  when  the  expansion  ratio  for  the  condition  of 
the  steam  corresponds  with  the  ratio  of  the  areas  of  the  mouth 
and  throat  of  the  nozzle.  A  little,  under-expansion  is  far  better, 
however,  than  the  same  amount  of  over-expansion,  meaning 
that  a  nozzle  that  is  too  small  for  the  required  expansion  is  more 

*  According  to  Dr.  0.  Recke,  if  the  total  divergence  of  a  nozzle  is  more  than 
6  degrees,  eddies  will  begin  to  form  hi  the  jet.  There  is  no  doubt  that  a  too  rapid 
divergence  produces  a  velocity  loss. 

f  Moyer's  Steam  Turbines. 


174  FLOW  OF  FLUIDS 

efficient  than  one  that  is  correspondingly  too  large.*  Fig.  62 
shows  a  curve  representing  average  values  of  nozzle  loss  f  used 
to  determine  discharge  velocities  from  nozzles  under  the  condi- 
tions of  under-  or  over-expansion. 

Non-expanding  Nozzles.  All  the  nozzles  of  Rateau  steam 
turbines  and  usually  also  those  of  the  low-pressure  stages  of  Curtis 
turbines  are  made  non-expanding;  meaning,  that  they  have  the 
same  area  at  the  throat  as  at  the  mouth.  For  such  conditions 
it  has  been  suggested  that  instead  of  a  series  of  separate  nozzles 
in  a  row  a  single  long  nozzle  might  be  used  of  which  the  sides 


•-i  10.19 


FIG.  63.  —  Non-expanding  Nozzles. 

were  arcs  of  circles  corresponding  to  the  inside  and  outside  pitch 
diameters  of  the  blades.  Advantages  would  be  secured  both  on 
account  of  cheapness  of  construction  and  because  a  largfc  amount 
of  friction  against  the  sides  of  nozzles  would  be  eliminated  by 
omitting  a  number  of  nozzle  walls.  Such  a  construction  has  not 
proved  desirable,  because  by  this  method  no  well-formed  jets 
are  secured  and  the  loss  from  eddies  is  excessive.  The  general 
statement  may  be  made  that  the  throat  of  a  well-designed  nozzle 
should  have  a  nearly  symmetrical  shape,  as  for  example  a  circle, 
a  square,  etc.,  rather  than  such  shapes  as  ellipses  and  long  rect- 

*  It  is  a  very  good  method  to  design  nozzles  so  that  at  the  rated  capacity  the 
nozzles  under-expand  at  least  10  per  cent,  and  maybe  20  per  cent.  The  loss  for 
these  conditions  is  insignificant,  and  the  nozzles  can  be  run  for  a  large  overload 
(with  increased  pressures)  in  nearly  all  types  without  immediately  reducing  the 
efficiency  very  much. 

t  C.  P.  Steinmetz,  Proc.  Am.  Soc.  Mech.  Engineers,  May,  1908,  page  628;  J. 
A.  Moyer,  Steam  Turbines-. 


MATERIALS  FOR  NOZZLES  175 

angles.  The  shape  of  the  mouth  is  not  important.  In  Curtis 
turbines  an  approximately  rectangular  mouth  is  used  because 
the  nozzles  are  placed  close  together  (usually  in  a  nozzle  plate 
like  Fig.  60)  in  order  to  produce  a  continuous  band  of  steam; 
and,  of  course,  by  using  a  section  that  is  rectangular  rather  than 
circular  or  elliptical,  a  band  of  steam  of  more  nearly  uniform 
velocity  and  density  is  secured. 

Fig.  63  shows  a  number  of  designs  of  non-expanding  nozzles 
used  by  Professor  Rateau.  The  length  of  such  nozzles  beyond 
the  throat  is  practically  negligible.  Curtis  non-expanding  noz- 
zles are  usually  made  the  same  length  as  if  expanding  and  the 
length  is  determined  by  the  throat  area. 

Materials  for  Nozzles.  Nozzles  for  saturated  or  slightly  super- 
heated steam  are  usually  made  of  bronze.  Gun  metal,  zinc 
alloys,  and  delta  metal  are  also  frequently  used.  All  these 
metals  have  unusual  resistance  for  erosion  or  corrosion  from  the 
use  of  wet  steam.  Because  of  this  property  as  well  as  for  the 
reason  that  they  are  easily  worked  with  hand  tools  *  they  are 
very  suitable  materials  for  the  manufacture  of  steam  turbine 
nozzles.  Superheated  steam,  however,  rapidly  erodes  all  these 
alloys  and  also  greatly  reduces  the  tensile  strength.  For  nozzles 
to  be  used  with  highly  superheated  steam,  cast  iron  is  generally 
used,  and  except  that  it  corrodes  so  readily  is  a  very  satisfactory 
material.  Commercial  copper  (about  98  per  cent)  is  said  to 
have  been  used  with  a  fair  degree  of  success  with  high  super- 
heats; but  for  such  conditions  its  tensile  strength  is  very  low. 
Steel  and  cupro-nickel  (8  Cu  +  2  Ni)  are  also  suitable  materials, 
and  the  latter  has  the  advantage  of  being  practically  non-cor- 
rodible. 

The  most  important  part  of  the  design  of  a  nozzle  is  the  deter- 
mination of  the  areas  of  the  various  sections  —  especially  the 
smallest  section,  if  the  nozzle  is  of  an  expanding  or  diverging 
type.  In  order  to  calculate  the  areas  of  nozzles  we  must  know 
how  to  determine  the  quantity  of  steam  (flow)  per  unit  of 
time  passing  through  a  unit  area.  It  is  very  essential  that  the 
*  Nozzles  of  irregular  shapes  are  usually  filed  by  hand  to  the  exact  size. 


176  FLOW  OF  FLUIDS 

nozzle  is  well  rounded  on  the  "  entrance  "  side  and  that  sharp 
edges  along  the  path  of  the  steam  are  avoided.  Otherwise  it 
is  not  important  whether  the  shape  of  the  section  is  circular, 
elliptical,  or  rectangular  with  rounded  corners. 

Whether  the  nozzle  section  is  throughout  circular,  square,  or 
rectangular  (if  these  last  sections  have  rounded  corners)  the 
efficiency  as  measured  by  the  velocity  will  be  about  96  to  97  per 
cent,  corresponding  to  an  equivalent  energy  efficiency  of  92  to  94 
per  cent. 

PROBLEMS 

1.  Air  at  a  temperature  of  100°  F.  and  pressure  of  100  Ibs.  per  sq.  in.  ab- 
solute flows  through  a  nozzle  against  a  back  pressure  of  20  Ibs.  per  sq.  in. 
absolute.     Assuming  the  initial  velocity  to  be  zero,  what  will  be  the  ve- 
locity of  discharge? 

2.  If  the  area  at  the  mouth  of  the  above  nozzle  is  0.0025  S(l-  ft-  and  the 
coefficient  of  discharge  is  unity,  how  many  pounds  of  air  will  be  discharged 
per  minute? 

3.  What  will  be  the  theoretical  kinetic  energy  per  minute  of  the  above 
jet  assuming  no  frictional  losses? 

4.  Steam  at  a  pressure  of  150  Ibs.  per  sq.  in.  absolute  and  3  per  cent 
moisture  flows  through  a  nozzle  against  a  back  pressure  of  17  Ibs.  per  sq.  in. 
absolute.     Calculate  the  velocity  at  the  throat  of  the  nozzle.     What  will 
be  the  velocity  at  the  end  of  the  nozzle? 

5.  Steam  at  a  pressure  of  200  Ibs.  per  sq.  in.  absolute,  temperature 
530°  F.,  expands  in  a  nozzle  to  a  vacuum  of  28.5  ins.  (30-in.  barometer). 
Calculate  the  absolute  velocity  of  the  steam  leaving  the  nozzle. 

6.  The  area  of  a  nozzle  at  its  smallest  section  is  0.75  sq.  in.     It  is  sup- 
plied with  dry  saturated  steam  at  160  Ibs.  per  sq.  in.  absolute  pressure  and 
the  exhaust  pressure  is  2  Ibs.  per  sq.  in.  absolute.    How  many  pounds  of 
steam  per  hour  will  the  nozzle  discharge? 

7.  A  steam  nozzle  is  to  be  designed  that  will  discharge  500  Ibs.  of  steam 
per  hour.    The  pressure  of  the  steam  is  175  Ibs.  per  sq.  in.  absolute  and  has 
100°  F.  superheat.    The  exhaust  pressure  is  28  ins.  vacuum  (barometer 
29.82).    What  will  be  the  area  of  the  nozzle  at  its  smallest  section  and  at  its 
end? 

8.  A  safety  valve  is  to  be  designed  for  a  200  horse  power  boiler  generating 
steam  at  150  Ibs.  per  sq.  in.  absolute  pressure  and  5  per  cent  moisture. 
Assuming  that  the  safety  valve  should  have  a  capacity  such  that  it  will 
release  the  boiler  of  all  steam  when  generating  double  its  rated  capacity, 
what  will  be  the  smallest  sectional  area  of  the  valve? 


PROBLEMS  177 

9.  An  injector  supplies  water  to  a  boiler.    The  boiler  pressure  is  100  Ibs. 
per  sq.  in.  absolute  and  the  steam  generated  is  dry  and  saturated,  the  tem- 
perature of  the  entering  feed  water  is  60°  F.  and  the  temperature  of  the  dis- 
charged mixture  of  condensed  steam  and  feed  water  is  180°  F.    The  suction 
head  is  3  ft.     What  is  the  weight  of  feed  water  supplied  to  the  boiler  per 
pound  of  steam?    What  is  the  mechanical  efficiency  of  the  injector? 

10.  Design  a  nozzle  to  deliver  400  Ibs.  of  steam  per  hour,  initial  pressure 
175  Ibs.  per  sq.  in.  absolute,  final  pressure  atmospheric  (barometer  28.62 
ins.),  temperature  of  steam  600°  F. 


CHAPTER  X 

APPLICATIONS  OF  THERMODYNAMICS  TO  COMPRESSED  AIR  AND 
REFRIGERATING   MACHINERY 

COMPRESSED  AIR 

Air  when  compressed  may  be  used  as  the  working  medium  in 
an  engine,  in  exactly  the  same  way  as  steam.  Furthermore,  it 
is  an  agent  for  the  transmission  of  power  and  can  be  distributed 
very  easily  from  a  central  station  for  the  purpose  of  driving 
engines,  operating  quarry  drills  and  various  other  pneumatic 
tools. 

Air  Compressors.  The  type  of  machine  used  for  the  compres- 
sion of  air  is  that  known  as  a  piston-compressor  and  consists  of 
a  cylinder  provided  with  valves  and  a  piston. 

The  work  performed  in  the  air-cylinder  of  a  compressor  can 
best  be  studied  successfully  from  an  indicator  diagram.  If  the 
compression  is  performed  very  slowly  in  a  conducting  cylinder, 
so  that  the  air  within  may  lose  heat  by  conduction  to  the  atmos- 
phere as  fast  as  heat  is  generated  by  compression,  the  process 
will  in  that  case  be  isothermal,  at  the  temperature  of  the  atmos- 
phere. Also  if  the  compressed  air  is  distributed  to  be  used  in 
compressed  air  motors  *  or  engines  without  a  change  of  temper- 
ature, and  that  the  process  of  expansion  in  the  compressed  air 
motors  or  engines  is  also  indefinitely  slow  and  consequently 
isothermal,  then  (if  we  neglect  the  losses  caused  by  friction  in 
pipes)  there  would  be  no  waste  of  power  in  the  whole  process  of 
transmission.  The  indicator  diagram  would  then  be  the  same 
per  pound  of  air  in  the  compressor  as  in  the  air  motor  or  engine, 
although  the  course  of  the  cycle  would  be  the  reverse  —  that  is, 
it  would  retrace  itself. 

*  Compressed  air  motors  are  similar  to  steam  engines,  but  use  compressed  air 
instead  of  steam. 

178 


WORK  OF   COMPRESSION 


179 


Adiabatic  compression  and  expansion  take  place  approximately 
if  the  expansion  and  compression  are  performed  very  quickly, 
or  when  the  air  is  not  cooled  during  compression,  —  in  such 
a  case  the  temperature  of  the  air  would  rise.  The  theoretical 
indicator  diagram  of  the  compressor,  Fig.  64,  is  FCBE  and  that 


Volume 
FIG.  64.  —  Diagram  of  Compressor. 


Volume 
FIG.  65.  —  Diagram  of  Air  Engine. 


of  the  air  engine,  Fig.  65,  is  EADF.  CB  and  AD  are  both 
adiabatic  lines.  The  change  of  volume  of  the  compressed  air 
from  that  of  EB  to  EA  occurs  through  its  cooling  in  the  dis- 
tributing pipes,  from  the  temperature  produced  by  adiabatic 
compression  down  to  the  temperature  of  the  atmosphere. 

Suppose  both  diagrams  of  the  compressor  and  of  the  air  engine 
are  superimposed  as  in  Fig.  66,  and  then  an  imaginary  isothermal 
line  is  drawn  between  the  points  A  and  C. 


Volume 
FIG.  66.  —  Superimposed  Diagrams  of  Figs.  21  and  22. 

It  is  then  evident  that  the  use  of  adiabatic  compression  causes 
a  waste  of  power  which  is  measured  by  the  area  ABC,  while  the 
use  of  the  adiabatic  expansion  in  the  air  engine  involves  a  further 
waste,  shown  by  the  area  ACD. 

Work  of  Compression.  Assuming  no  clearance  in  the  com- 
pressor and  isothermal  compression,  the  pressure- volume  diagram 
will  be  similar  to  Fig.  64.  The  work  of  the  cycle  will  be 


l8o    COMPRESSED  AIR  AND  REFRIGERATING  MACHINERY 


W   =  PcVc  -  PcVc  loge  £   -  PbVb,  (236) 

*  b 

which  becomes,  since  PCVC  =  P&F&, 

W  =  -  PcVc  loge  Y  =  PcVc  loge  ^-  (237) 

In  practice  the  compression  cannot  be  made  strictly  isother- 
mal, as  the  operation  of  the  piston  would  be  too  slow.  The  dif- 
ference between  isothermal  and  adiabatic  compression  (and  ex- 
pansion) can  be  shown  graphically  as  in  Figs.  67  and  68.  In 

c    B 


Volume 
FIG.  67.  —  Compression  Diagram. 


Volume 
FIG.  68.  —  Expansion  Diagram. 


these  illustrations  the  terminal  points  are  correctly  placed  for  a 
certain  ratio  for  both  compression  and  expansion.  Note  that  in 
the  compressing  diagram  (Fig.  67)  the  area  between  the  two 
curves,  ABC,  represents  the  work  lost  in  compressing  due  to  heat- 
ing, and  the  area  between  the  two  curves,  ACMNF  (in  Fig.  68), 
shows  the  work  lost  by  cooling  during  the  expansion.  The 
isothermal  curve  AC  will  be  the  same  for  both  cases. 

The  temperature  of  the  air  is  prevented  as  far  as  possible 
from  rising  during  trie  compression  by  injecting  water  into  the 
compressing  cylinder,  and  in  this  way  the  compression  curve  will 
change.  The  curves  which  would  have  been  PV  =  a  constant,  if 
isothermal,  and  PVl4  =  a,  constant,  if  adiabatic,  will  be  very  much 
modified.  In  perfectly  adiabatic  conditions  the  exponent  " n" 
=  i  .40  for  air,  but  in  practice  the  compressor  cylinders  are  water- 
jacketed,  and  thereby  part  of  the  heat  of  compression  is  con- 
ducted away,  so  that  " n"  becomes  less  than  1.40.  This  value 
of  " n"  varies  with  conditions;  generally  the  value  is  between 
1.2  and  1.3. 


EFFECT  OF   CLEARANCE  UPON   VOLUMETRIC  EFFICIENCY      181 

When  the  compression  curve  follows  the  law,  PVn  equals  a 
constant,  work  of  compression  (Fig.  64)  is 

W  =  PcVc  +  PcVc  ~  P»V»  -  PbVb 
n  —  i 

n     (PCVC  -  P.7.),  (238) 


n  —  i 

i 


since  PcVcn  =  PbVbn;  Vb  =  ^  (239) 

Combining  equations  (238)  and  (239), 


The  Effect  of  Clearance  upon  Volumetric  Efficiency.     It  is 

impossible  to  construct  a  compressor  without  clearance,  con- 
sequently the  indicator  diagram  differs  from  the  ideal  case.  At 
the  end  of  the  discharge  stroke,  the  clearance  volume  is  filled  with 
compressed  air.  When  the  piston  moves  on  its  outward  stroke, 
the  clearance  air  expands  and  the  suction  valves  of  the  compres- 
sor will  be  held  shut  until  the  piston  has  moved  a  sufficient  dis- 
tance to  permit  the  entrapped  air  to  expand  to  atmospheric 
pressure.  When  that  point  is  reached  any  further  movement  of 
the  piston  opens  the  suction  valves  and  external  air  is  drawn  into 
the  cylinder.  Thus  the  entire  stroke  of  the  compressor  piston 
is  not  effective  in  pumping  air.  The  ratio  of  the  volume  of  air 
pumped  to  the  volume  swept  by  the  piston,  or  piston  displace- 
ment of  the  cylinder,  is  termed  volumetric  efficiency. 

Fig.  69  illustrates  an  ideal  compressor  diagram  with  clearance. 
The  air  entrapped  in  the  clearance  space  equals 

fls  =  CVS, 

where  Vs  =  volume  swept  or  piston  displacement  of  the  cylinder. 
C  =  percentage  of  clearance." 

The  air  in  Fig.  69  expands  to  F4  at  which  point  the  inlet  valves 
open.     The  air  drawn  into  the  cylinder  is  represented  by  the 


182    COMPRESSED  AIR  AND  REFRIGERATING  MACHINERY 


difference  in  volume  between  V\  and  F4.  This  air,  as  well  as  the 
clearance  air,  is  compressed  to  point  2,  while  the  compressed  air 
is  discharged  from  points  2  to  3.  Knowing  the  per  cent  of  clear- 


volume 

FIG.  69.  —  Ideal  Air  Compressor  with  Clearance. 

ance,  the  volume  swept  by  the  piston  (Fs),  and  the  initial  and 
final  pressure,  the  volumetric  efficiency  (Fer)  may  be  deter- 
mined from  the  following  equations: 

-  F4 


Fef.    = 

P,  (cvs)n  = 


vs 

P4F4n. 


since 


=  F8  +  CV, 


V1  -  Vt  =  V,  +  CV,  -  CVS 


TWO  STAGE   COMPRESSION 

Therefore  volumetric  efficiency  is 


(240 


The  volumetric  efficiency  also  decreases  with  the  altitude,  as 
the  weight  of  a  cubic  foot  of  air  decreases  as  the  altitude  increases. 

Two  Stage  Compression.  The  problem  of  economy,  obviously, 
becomes  one  of  abstracting  the  heat  generated  in  the  air  during 
the  process  of  compression.  As  previously  mentioned,  this  is 
partially  accomplished  by  water-jacketing  the  cylinders,  and  also 
by  water  injection.  Nevertheless,  owing  to  the  short  interval 
within  which  the  compression  takes  place,  and  the  comparatively 
small  volume  of  air  actually  in  contact  with  the  cylinder  walls, 
very  little  cooling  really  occurs.  The  practical  impossibility  of 
proper  cooling  to  prevent  waste  of  energy  leads  to  the  alternative 
of  discharging  air  from  one  cylinder  after  partial  compression  has 
been  effected,  into  a  so-called  inter-cooler,  intended  to  absorb 
the  heat  generated  during  the  first  compression,  and  then 
compressing  the  air  to  the  final  pressure  in  another  cylinder. 


Volume 
FIG.  70.  —  Indicator  Diagram  of  Two-stage  Air  Compressor. 

This  operation  is  termed  "  two-stage  "  compression  and  when 
repeated  one  or  more  times  for  high  pressures  the  term  "  multi- 
stage "  compression  applies. 

Referring  to  Fig.  70  and  assuming  the  compression  in  a  two- 
stage  compressor  to  be  adiabatic  for  each  cylinder,  the  compres- 


184     COMPRESSED  AIR  AND  REFRIGERATING  MACHINERY 

sion  curve  is  represented  by  the  broken  line  ABDE  ;  the  compres- 
sion proceeds  adiabatically  in  the  first  or  low-pressure  cylinder 
to  B;  the  air  is  then  taken  to  a  cooler  and  cooled  under  practi- 
cally constant  pressure  until  its  initial  temperature  is  almost 
reached,  and  its  volume  reduced  from  HB  to  HD;  it  is  then 
introduced  to  the  second  or  high-pressure  cylinder  and  com- 
pressed adiabatically  along  the  line  DE  to  the  final  pressure 
condition  that  was  desired.  It  is  seen  that  the  compression 
curve  approaches  the  isothermal  line  FA.*  The  isothermal  con- 
dition is  obviously  desired  and,  in  consequence,  air-machines 
are  built  to  approach  that  condition  as  nearly  as  possible. 

Referring  to  Fig.  70,  the  work  of  each  stage  from  equation  (240) 
will  be 

stage)  =  P 


/  »-K 

W  (ad  stage)  =  -*—PtVL  -  (£  Y   )• 
n  —  i          \        \*a/      ' 


The  total  work  of  compression  is 

W  (total)  =  W  (ist  stage)  +  W  fcd  stage) 


n— 1 


With  perfect  cooling  PaVa  =  PaV^  also  P*  =  Pd\  Pc  =  PC,  then 


The  work  of  compression  as  expressed  by  equation  (242)  be- 
comes a  minimum  when 

n-l  n-1 


is  a  maximum.     Since  the  initial  and  final  pressures  Pa  and  Pc 
are  fixed,  the  pressure  in  the  receiver,  Pb,  can  be  found  by  differ- 

*  The  line  FE  represents  further  cooling. 


REFRIGERATING  MACHINES  OR  HEAT  PUMPS  18$ 

entiating  equation  (243)  and  equating  this  differential  to  zero,  or 

dp 


(244) 


For  compressing  air  to  high  pressures  three  and  four  stage  com- 
pressors are  used. 

REFRIGERATING   MACHINERY 

Refrigerating  Machines  or  Heat  Pumps.  By  a  refrigerating 
machine  or  heat  pump  is  meant  a  machine  which  will  carry  heat 
from  a  cold  to  a  hotter  body.*  This,  as  the  second  law  of 
thermodynamics  asserts,  cannot  be  done  by  a  self-acting  proc- 
ess, but  it  can  be  done  by  the  expenditure  of  mechanical  work. 
Any  heat  engine  will  serve  as  a  refrigerating  machine  if  it  be  forced 
to  trace  its  indicator  diagram  backward,  so  that  the  area  of  the 


Volume 
FIG.  71.  —  Pressure- volume  Diagram  of  Carnot  Cycle. 

diagram  represents  work  spent  on,  instead  of  done  by,  the 
working  substance.  Heat  is  then  taken  in  from  the  cold  body 
and  heat  is  rejected  to  the  hot  body. 

*  This  statement  is  not  at  variance  with  our  knowledge  that  heat  does  not 
flow  of  itself  from  a  cold  body  to  a  hotter  body. 


l86    COMPRESSED  AIR  AND  REFRIGERATING  MACHINERY 

Take  the  Carnot  cycle,  using  air  as  working  substance  (Fig. 
71),  and  let  the  cycle  be  performed  in  the  order  dcba,  so  that  the 
area  of  the  diagram  is  negative,  and  represents  work  spent  upon 
the  working  substance.  In  the  stage  dc,  which  is  isothermal 
expansion  in  contact  with  the  cold  body  R  (as  in  Fig.  7,  page  41), 
the  air  takes  hi  a  quantity  of  heat  from  R  equal  to  wRT^,  log«  r 
[equation  (46)],  and  in  stage  ba  it  gives  out  to  the  hot  body  H  a 
quantity  of  heat  equal  to  wRT\  log«  r.  There  is  no  transfer  of 
heat  in  stages  cb  and  ad.  Thus  R,  the  cold  body,  is  constantly 
being  drawn  upon  for  heat  and  can  therefore  be  maintained  at  a 
temperature  lower  than  its  surroundings.  In  an  actual  refriger- 
ating machine  operating  with  air,  the  cold  body  R  consists  of  a 
coil  of  pipe  through  which  brine  circulates  while  "  working  "  air 
is  brought  into  contact  with  the  outside  of  the  pipe.  The  brine 
is  kept,  by  the  action  of  the  machine,  at  a  temperature  below 
32°  F.  and  is  used  in  its  turn  to  extract  heat  by  conduction  from 
the  water  which  is  to  be  frozen  to  make  ice.  The  "  cooler  "  H, 
which  is  the  relatively  hot  body,  is  kept  at  as  low  a  temperature 
as  possible  by  means  of  circulating  water,  which  absorbs  the  heat 
rejected  to  H  by  the  "  working  "  air. 

The  size  of  an  air  refrigerating  machine  is  very  large  as  com- 
pared with  its  performance.  The  use  of  a  regenerator,  as  in 
Stirling's  engine  (Fig.  8),  may  be  resorted  to  in  place  of  the 
two  adiabatic  stages  in  the  Carnot  cycle,  with  the  advantage  of 
making  the  machine  much  less  bulky.  Refrigerating  machines 
using  air  as  working  substance,  with  a  regenerator,  were  intro- 
duced by  Dr.  A.  C.  Kirk  and  have  been  widely  used.*  The 
working  air  is  completely  enclosed,  which  allows  it  to  be 
in  a  compressed  state  throughout,  so  that  even  its  lowest 
pressure  is  much  above  that  of  the  atmosphere.  This  makes 
a  greater  mass  of  air  pass  through  the  cycle  in  each  revolution 
of  the  machine,  and  hence  increases  the  performance  of  a 
machine  of  given  size.  In  all  air  refrigerating  machines  the 

*  See  Kirk,  On  the  Mechanical  Production  of  Cold,  Proc.  Inst.  of  C.  £.,  vol. 
XXXVII,  1874.  Also  lectures  on  Heat  and  its  Mechanical  Applications,  in  the 
same  proceedings  for  1884. 


SYSTEMS  OF  MECHANICAL  REFRIGERATION  187 

temperature  range  must  be  high  to  produce  a  given  refrigerating 
effect. 

In  another  class  of  refrigerating  machines  the  working  sub- 
stance, instead  of  being  air;  consists  of  a  liquid  and  its  vapor, 
and  the  action  proceeds  by  alternate  evaporation  under  a  low 
pressure  and  condensation  under  a  relatively  high  pressure.  A 
liquid  must  be  chosen  which  evaporates  at  the  lower  extreme  of 
temperature  under  a  pressure  which  is  not  so  low  as  to  make 
the  bulk  of  the  engine  excessive.  Ammonia,  ether,  sulphurous 
acid,  and  other  volatile  liquids  have  been  used.  Ether  machines 
are  inconveniently  bulky  and  cannot  be  used  to  produce  intense 
cold,  for  the  pressure  of  that  vapor  is  only  about  1.3  pounds  per 
square  inch  at  4°  F.,  and  to  make  it  evaporate  at  any  tempera- 
ture nearly  as  low  as  this  would  require  the  cylinder  to  be  exces- 
sively large  in  proportion  to  the  performance.  This  would  not 
only  make  the  machine  clumsy  and  costly,  but  would  involve 
much-  waste  of  power  in  mechanical  friction.  The  tendency  of 
the  air  outside  to  leak  into  the  machine  is  another  practical  ob- 
jection to  the  use  of  so  low  a  pressure.  With  ammonia  a  dis- 
tinctly lower  limit  of  temperature  is  practicable:  the  pressures 
are  rather  high  and  the  apparatus  is  compact.  Carbo.nic  acid 
has  been  used  as  a  refrigerant  in  small  machines.  The  objection 
to  carbonic  acid  is  that  the  pressures  are  very  high  as  compared 
with  ammonia  (see  Appendix).  The  critical  temperature  of 
carbonic  acid  is  less  than  90°  F. 

Unit  of  Refrigeration.  The  capacity  of  a  refrigerating  ma- 
chine is  usually  expressed  in  .tons  of  refrigeration  or  ice  melting 
effect  per  twenty-four  hours.  As  the  latent  heat  of  fusion  of 
ice  is  about  144  B.t.u.,  the  heat  units  withdrawn  per  ton  of  re- 
frigerating effect  per  twenty-four  hours  is 

2000  X  144  =  288,000  B.t.u. 

Systems  of  Mechanical  Refrigeration.  The  standard  systems 
of  mechanical  refrigeration  are:* 

(A)  The  dense-air  system,  so-called  because  the  air  which  is 

*  Lucke's  Engineering  Thermodynamics,  page  1148. 


1 88     COMPRESSED  AIR  AND  REFRIGERATING  MACHINERY 


the  medium  is  never  allowed  to  fall  to  atmospheric  pressure,  so 
as  to  reduce  the  size  of  the  cylinders  and  pipes  through  which  a 
given  weight  is  circulating. 

(B)  The  compression  system,  using  ammonia,  carbon  dioxide 
or  sulphur  dioxide,  and  so-called  to  distinguish  it  from  the  third 


Circulating 
Water 


Cool  Com- 
pressed Air 


Water  Cooler 


Hot  Corn- 
Air 


Cold  Low 
Pressure  Air 


^,  Brine  Tank 


Warm 
Brine 


'Circulating  Brine  I 


Warmed  Low 
Pressure  Air 


Pipes 


/Cold 
Brine 


FIG.  72.  —  Dense  Air  System  of  Refrigeration. 

system,  because  a  compressor  is  used  to  raise  the  pressure  of  the 
vapor  and  deliver  it  to  the  condenser  after  removing  it  from  the 
evaporator. 
'  (C)  The  absorption  system,  using  ammonia,  and  so-called  be- 


THE  AIR   SYSTEM   OF  REFRIGERATION  189 

cause  a  weak  water  solution  removes  vapor  from  the  evaporator 
by  absorption,  the  richer  aqua  ammonia  so  formed  being  pumped 
into  a  high-pressure  chamber  called  a  generator  in  communica- 
tion with  the  condenser,  where  the  ammonia  is  discharged  from 
the  liquid  solution  to  the  condenser  by  heating  the  generator, 
to  which  the  solution  is  delivered  by  the  pump. 

No  matter  what  system  is  used,  circulating  water  is  employed 
to  receive  the  heat,  the  temperature  of  which  limits  the  highest 
temperature  allowable  in  the  system  and  indirectly  the  highest 
pressure. 

The  Air  System  of  Refrigeration.  The  dense  or  closed  air 
system  is  illustrated  in  Fig.  72,  in  which  air,  previously  freed  of 
moisture,  is  continuously  circulated.  The  engine  cylinder  E 
furnishes  power  *  to  drive  the  compressor  cylinder  F.  This 
cylinder  delivers  hot-compressed  air  into  a  cooler  A  where  it 
is  cooled,  and  then  passed  on  to  the  expansion  cylinder  G  (tan- 
dem-connected to  both,  the  compressor  F  and  to  the  engine 
cylinder  E),  which  in  turn  sends  cold  low-pressure  air  first  through 
the  refrigerating  coils  in  the  brine  tank 
B  and  then  back  to  the  compressor  cylin- 
der F;  thus  the  air  cycle  is  completed.  | 
The  courses  of  the  circulating  water  and 
also  of  the  brine  are  shown  by  the  dotted 
lines. 

The  dense-air  cycle  in  a  pressure- volume  FIG.  73 .  —  Pressure-vol- 
diagram  is  represented  in  Fig.  73,  in  which  ume  Diagram  of  Dense  Air 
BC  is  the  delivered  volume  of  hot-corn-  Cycle 
pressed  air;  CM  is  the  volume  of  cooled  air  admitted  in  the  ex- 
pansion cylinder;  MB  the  reduction  in  volume  due  to  the  water 
cooler;  MN  the  expansion;  NA  the  refrigeration  or  heating  of 
the  air  by  the  brine,  and  AB  the  compression.  This  operation  is 
but  a  reproduction  of  that  previously  described. 

The  work  Wc  expended  in  the  compression  cylinder  F  is  DABC 

*  Since  the  work  done  by  the  expansion  of  the  cool-compressed  air  is  less  than 
that  necessary  for  the  compressing  of  the  air  taken  from  the  brine  coils  through 
the  same  pressure  conditions,  a  means  must  be  employed  to  make  up  for  the  dif- 
ference, and  for  this  purpose  the  engine  cylinder  is  used. 


I  go    COMPRESSED  AIR  AND  REFRIGERATING  MACHINERY 

(Fig.  73)  ;  that  done  by  the  expansion  cylinder  G  is  We  =  DCMN. 
The  shaded  area  MEAN  represents  the  work  which  must  be  sup- 
plied by  the  engine  E. 

If  w  pounds  of  air  are  passing  through  the  refrigerating  machine 
per  minute,  the  heat  withdrawn  from  the  cold  room  or  absorbed 
by  the  brine  along  NA  (Fig.  73)  is 

QNA  =  wCp  (ta  —  /»).  (245) 

The  work  expended  in  compressing  w  pounds  of  air,  assuming 
polytropic  compression,  is,  by  equation  (240), 


If  the  compression  is  adiabatic, 

(246) 


since  ~  =  ^  also  AR  =  Cp-  C,  =  C 


If  the  expansion  is  complete  in  the  expansion  cylinder,  the  work 
done  in  expansion  is 

PT.-^(A.-A,).      '  ".;.       (247) 

The  net  work  of  the  engine  E  required  to  produce  refrigeration 
becomes 

'    W   =   We   -  We   =  ^   (fo   -  ta   ~  tm  +  *»).      (248) 

A 
Since,  from  equation  (245), 

w  =     ^  (ta  —  /„),  equation  (248)  becomes 


THE  VAPOR   COMPRESSION  SYSTEM  OF  REFRIGERATION      191 


The  Vapor  Compression  System  of  Refrigeration.  The  com- 
pression system  for  ammonia  or  similar  condensable  vapors  is 
shown  in  Fig.  74.  The  figure  illustrates  the  essential  mem- 
bers of  a  complete  compression  refrigerating  system.  B  repre- 


Circulating 
V  Water 

I 


High  Pres-^ 
sure  Liquid 


W 
Condenser 


High  Pres- 
sure Vapor 


Throttling  or 
Expansion  Valve 

Low  Pressure  Liquid 


Brine  ^    Tank 


Low  Pres- 
sure Vapor 


It — Circulating  Brine — >1 
Pipes  | 

Warm  f  \  Cold 

Brine  I  |  Brine 

FIG.  74.  —  Compression  System  of  Refrigeration. 

sents  the  direct-expansion  coil  in  which  the  working  medium  is 
evaporated;  F,  the  compressor  or  pump,  for  increasing  the  pres- 
sure of  the  gasified  ammonia;  E,  the  engine  cylinder,  —  the  source 
of  power;  W,  the  condenser,  for  cooling  and  liquefying  the  gasi- 
fied ammonia;  and  V,  a  throttling  valve,  by  which  the  flow  of 


COMPRESSED  AIR  AND  REFRIGERATING  MACHINERY 

liquefied  ammonia  under  the  condenser  pressure  is  controlled  as 
it  flows  from  the  receiver  R  to  the  expansion  coils;  B,  the  brine 
tank,  in  which  a  materially  lower  pressure  is  maintained  by  the 
pump  or  compressor  in  order  that  the  working  medium  may  boil 
at  a  sufficiently  low  temperature  to  take  heat  from  and  con- 
sequently refrigerate  the  brine  which  is  already  cooled. 

The  operation  of  the  compressor  F  (Fig.  74)  is  theoretically  a 
reversed  Rankine  cycle  (Figs.  31  and  32).  If  one  pound  of  vapor 
is  passed  through  the  system,  the  work  (Wc)  of  the  compressor  is 

Wc  =  (Hz  -  Hi)  (B.t.u.), 
where 

Hi  =  the  total  heat  of  the  vapor  at  entrance  to  the  com- 

pressor, 
Hz  =  the  total  heat  of  the  vapor  discharged  from  the  com- 

pressor. 

The  vapor  entering  the  compressor  may  be  treated  as  though 
it  were  dry  and  saturated.  The  entropy-temperature  diagram 
(Fig.  75)  illustrates  this  case.  The  vapor  discharged  by  the 
compressor  will  then  be  superheated,  having  a  temperature 
ts  at  a  pressure  PZ.  The  condition  of  the  discharged  vapor  is 
determined  by  equating  the  entropies  at  the  inlet  and  discharge 
pressures.  The  total  heat  (Hz)  of  the  discharged  vapor  will  be 

Hz  =  [hz  +  Lz  +  CP  (ts  -  tz)].  (250) 

The  horse  power  of  the  compressor  is,  if  w  pounds  of  vapor  is 
circulated  per  minute: 


The  heat  absorbed  per  minute  by  w  pounds  of  vapor  passing 
through  the  refrigerating  or  brine  coils  will  be 

Qr  =  w  (^  +  Li-ho) 

=  W(H!-AO).  (252) 

ho  =  the  liquid  heat  of  the  vapor  after  leaving 
the  condenser. 


THE   VAPOR  ABSORPTION   SYSTEM   OF   REFRIGERATION      193 


The  heat  absorbed  per  minute  by  the  condenser  (Qc)  is 
w  (H2  -  ho)  (B.t.u.). 


(253) 


p, 


The  heat  absorbed  by  the  condenser  is  theoretically  equal  to 
the  sum  of  the  heat  absorbed  in  the  refrigerator  and  the  heat 
equivalent  of  the  work  of  compression. 

Ammonia  compressors  are  operated  either  on  the  dry  or  on 
the  wet  system.  In  the  dry  system  the  ammonia  entering  the 
compressor  is  a  dry  vapor  as  illustrated  in  Fig.  75.  In  the  wet 
system  the  ammonia  enters  the  compressor  in  the  wet  state,  the 
heat  developed  during  the  com- 
pression being  used  in  evapo- 
rating the  liquid  ammonia  into 
a  vapor. 

The  Vapor  Absorption  Sys- 
tem of  Refrigeration.  The  ab-  i 
sorption  system  depends  upon 
the  fact  that  anhydrous  am- 
monia possesses  the  property 
of  forming  aqua  ammonia. 
The  amount  of  ammonia  water 
will  absorb  depends  upon  the 
temperature  of  the  water;  the 
colder  the  water,  the  greater  are 
its  absorptive  powers.  Entropy -0 

The  absorption  system  differs  FlG-  75-  —  Entropy-Temperature  Dia- 
e  i  •  cram  for  Ammonia 

from    the   vapor    compression 

system  in  that  the  absorption  system  replaces  the  compressor 
by  an  absorber,  where  the  anhydrous  ammonia  is  changed  into 
aqua  ammonia,  a  pump  which  transfers  the  ammonia  from  the 
absorber  to  the  generator,  and  a  generator,  where  the  aqua 
ammonia  is  heated.  Both  systems  have  a  condenser  where  the 
ammonia  is  cooled  and  liquefied  and  an  expansion  valve  or 
throttling  valve  by  means  of  which  the  flow  of  liquid  ammonia 
to  the  expansion  coils  is  controlled.  In  the  absorption  system 
the  anhydrous  ammonia  vapor  flows  from  the  expansion  coils  to 


IQ4    COMPRESSED  AIR  AND  REFRIGERATING  MACHINERY 

the  absorber,  in  which  the  anhydrous  vapor  comes  in  contact 
with  weak  aqua  ammonia.  The  weak  ammonia  absorbs  the 
ammonia  vapor.  From  the  absorber  the  ammonia  is  pumped 
into  the  generator,  where  it  is  heated  by  steam  coils.  The  vapor 
driven  off  in  the  generator  passes  to  the  condenser  and  from  there 
through  the  expansion  or  throttle  valve  to  the  expansion  coils 
which  are  located  in  the  brine  tank  or  in  the  refrigerating  room. 
The  absorption  system  is  usually  provided  with  a  rectifier  to 
thoroughly  dry  the  gas  before  it  enters  the  condenser,  an  ex- 
changer which  heats  the  strong  ammonia  by  cooling  the  weak 
ammonia,  and  with  other  auxiliary  equipment  to  reduce  the  heat 
losses  in  the  system. 

COEFFICIENT    OF    PERFORMANCE    OF    REFRIGERATING    MACHINES 

The  Coefficient  of  Per- 1  _  Heat  extracted  from  the  cold  body 
formance  Work  expended 

This  ratio  may  be  taken  as  a  coefficient  of  performance  in  esti- 
mating the  merits  of  a  refrigerating  machine.  When  the  limits 
of  temperature  7\  and  T%  are  assigned  it  is  very  easy  to  show  by 
a  slight  variation  of  the  argument  used  in  Chapter  IV  that  no 
refrigerating  machine  can  have  a  higher  coefficient  of  performance 
than  one  which  is  reversible  according  to  the  Carnot  method. 
For  let  a  refrigerating  machine  S  be  driven  by  another  R  which 
is  reversible  and  is  used  as  a  heat-engine  in  driving  S.  Then  if  S 
had  a  higher  coefficient  of  performance  than  R  it  would  take  from 
the  cold  body  more  heat  than  R  (working  reversed)  rejects  to 
the  cold  body,  and  hence  the  double  machine,  although  purely 
self-acting,  would  go  on  extracting  heat  from  the  cold  body  in 
violation  of  the  Second  Law  of  Thermodynamics.  Reversi- 
bility, then,  is  the  test  of  perfection  in  a  refrigerating  machine 
just  as  it  is  in  a  heat-engine. 

When  a  reversible  refrigerating  machine  takes  in  all  its  heat, 
namely  Qc  at  TZj  and  rejects  all,  namely  Qa  at  T\,  and  if  we  repre- 
sent the  heat  equivalent  of  the  work  done  by  W  =  Qa  —  Qc,  then 
the  coefficient  of  performance  is,  as  already  defined, 


PROBLEMS  195 

&-          &  T*  (ocA 

W         Qa-Qc          7\  -  Tz' 

Hence,  the  smaller  the  range  of  temperature,  the  better  is  the 
performance.  To  cool  a  large  mass  of  any  substance  through  a 
few  degrees  will  require  much  less  expenditure  of  energy  than  to 
cool  one-fifth  of  the  mass  through  five  times  as  many  degrees, 
although  the  amount  of  heat  extracted  is  the  same  in  both 
cases.  If  we  wish  to  cool  a  large  quantity  of  a  substance  it  is 
better  to  do  this  by  the  direct  action  of  a  refrigerating  machine 
working  through  the  desired  range  of  temperature,  than  to  cool 
a  portion  through  a  wider  range  and  then  let  this  mix  with  the 
rest.  This  is  only  another  instance  of  a  wide,  general  principle^ 
that  any  mixture  of  substances  at  different  temperatures  is  ther- 
modynamically  wasteful  because  the  interchange  of  heat  between 
them  is  irreversible.  An  ice-making  machine,  for  example, 
should  have  its  lower  limit  of  temperature  only  so  much  lower 
than  32°  F.  as  will  allow  heat  to  be  conducted  to  the  working 
fluid  with  sufficient  rapidity  from  the  water  that  is  to  be  frozen. 

PROBLEMS 

1.  If  200  gu.  ft.  of  free  air  per  minute  (sea-level)  is  compressed  isother- 
mally  and  then  delivered  into  a  receiver,  the  internal  pressure  of  which  is 
102.9  Ibs.  per  sq.  in.  absolute,  find  the  theoretical  horse  power  required. 

2.  What  will  be  the  net  work  done  in  foot-pounds  per  stroke  by  an  air 
compressor  displacing  3  cu.  ft.  per  stroke  and  compressing  air  from  atmos- 
pheric pressure  to  a  gage  pressure  of  75  Ibs.?     (Isothermal  compression.) 

3.  WTiat  horsepower  will  be  needed  to  compress  adiabatically  1500 
cu.  ft.  of  free  air  per  minute  to  a  gage  pressure  of  58.8  Ibs.,  when  n  equals 
1.4? 

4.  A  compressed-air  motor  without  clearance  takes  air  at  a  condition 
of  200  Ibs.  per  sq.  in.  (gage)  and  operates  under  a  cut-off  at  one-fourth 
stroke.    What  is  the  work  in  foot-pounds  that  can  be  obtained  per  cubic 
foot  of  compressed  air,  assuming  free  air  pressure  of  14.5  Ibs.  and  n  equal 
to  1.41? 

5.  Find  the  theoretical  horse  power  developed  by  3  cu.  ft.  of  air  per 
minute  having  a  pressure  of  200  Ibs.  per  sq.  in.  absolute,  if  it  is  admitted 
and  expanded  in  an  air  engine  with  one-fourth  cut-off.    The  value  of  n  is 
1.2.    (Neglect  clearance.) 


196      COMPRESSED  AIR  AND  REFRIGERATING  MACHINERY 

6.  Compute  the  net  saving  in  energy  that  is  effected  by  compressing 
isothermally  instead  of  adiabatically  50  cu.  ft.  of  free  air  to  a  pressure  of 
200  Ibs.  per  sq.  in.  gage.     Barometer  =  14  Ibs.  per  sq.  in.,  and  a  tempera- 
ture of  70°  F.    What  is  the  increase  in  intrinsic  energy  during  each  kind 
of  compression?    How  much  heat  is  lost  to  the  jacket- water  during  each 
kind  of  compression? 

7.  Let  a  volume  of  12  cu.  ft.  of  free  air  be  adiabatically  compressed  in  a 
one  stage  air  compressor  from  atmospheric  pressure  (15  Ibs.)  to  85  Ibs.  gage; 
the  initial  temperature  of  the  air  being  70°  F. 

(a)  What  is  the  volume  and  temperature  of  the  air  after  the  com- 

pression? 

(b)  Suppose  this  heated  and  compressed  air  is  cooled  to  an  initial 

temperature  of  60°  F.  at  constant  volume,  what  is  its  pressure 
for  that  condition? 

(c)  Now  if  the  air  occupies  such  a  volume  as  found  in  (a)  and  at 

an  absolute  pressure  as  determined  in  (b),  at  a  temperature 
of  60°  F.,  and  is  then  allowed  to  expand  adiabatically  down 
to  atmospheric  pressure  (15  Ibs.),  what  is  the  temperature 
and  volume  of  the  expanded  air  in  Fahrenheit  degrees? 

8.  Compare  the  work  required  to  compress  one  pound  of  air  from  atmos- 
pheric pressure  and  a  temperature  of  60°  F.  to  a  pressure  of  200  Ibs.  per  sq. 
in.  absolute  in  (i)  a  one-stage  and  (2)  in  a  two-stage  compressor.    Assume 
n  =  1.25. 

9.  Using  equation  249  as  a  basis,  deduce  a  formula  for  the  horse  power 
required  to  abstract  a  given  number  of  heat  units  per  minute  by  an  air 
refrigerating  machine. 

10.  Calculate  the  approximate  dimensions  of  a  ten-ton  air  refrigerating 
machine  and  the  power  required  to  drive  it.     Pressure  in  the  cold  chamber 
is  atmospheric  and  the  temperature  32°  F.     Pressure  of  air  delivered  by 
compressor  is  90  Ibs.  per  sq.  in.  gage.    The  temperature  of  the  air  coming 
from  condenser  is  85°  F.  and  the  machine  operates  at  60  r.p.m.    Allow 
12  Ibs.  drop  in  pressure  between  the  compressor  and  the  expanding  cylinder. 

11.  Calculate  the  approximate  dimensions  of  a  ten-ton  ammonia  com- 
pression refrigerating  machine  and  the  power  required  to  drive  it.    The 
temperature  in  the  expansion  coils  is  15°  F.  and  the  temperature  in  the  con- 
denser is  85°  F.    The  machine  is  double  acting  and  operates  at  a  speed  of 
80  r.p.m. 

12.  What  is  the  ice-making  capacity  per  twenty-four  hours  of  the  machine 
in  problem  1 1?    The  temperature  of  the  water  to  be  frozen  is  80°  F. 


APPENDIX 


TABLE  I 
SPECIFIC  HEAT  OF  GASES  AND  VAPORS 

(Taken  from  Smithsonian  Physical  Tables) 


Substance 

Range  of 
Temp.0  C. 

Cp 

Authority 

Cp 

Cv 

Authority 

Calcu- 
lated 
Cv 

—  30-10 

0.23771 

Regnault 

o-ioo 

0.23741 

" 

Air  

--.— 

^ 

ii 

O     2W 

°-23751 

20-IOO 

0.2389 

Wiedemann 

Mean 

0.23788 

1.4066 

Various 

0.1691 

Alcohol, 
(ethyl) 

108-220 

0.4534 

Regnault 

1.136 

l     Jaeger,     1 
1  Neyreneuf  / 

0.3991 

Alcohol, 
(methyl) 

101-223 

0.4580 

Regnault 

23-100 

0.5202 

Wiedemann 

27-200 

0.5356 

(i 

Ammonia.  .  . 

24-216 

0.5125 

Regnault 

i                      \ 

Mean 

0.5228 

I    31 

Cazin, 

TTT--11 

0.3991 

•"•  •Ox 

(    Wullner     J 

34-"5 

o  .  2990 

Wiedemann 

Benzene  

35-i8o 

0.3325 

11 

116-218 

0.3754 

Regnault 

-28-77 

0.1843 

Regnault 

Carbon 

15-100 

0.2025 

" 

Dioxide 

11-214 

0.2169 

ii 

(•v 

Mean 

O.2OI2 

I    3OO 

Wullner    } 

0.1548 

A    .  ^  W 

Rontgen,    J 

Carbon 

23-99 

0.2425 

Wiedemann 

Monoxide 

26-198 

0.2426 

» 

I-403 

f      Cazin,      I 
I    Wullner    j 

0.1729 

69-224 

0-4797 

Regnault 

Ether  

27-189 

0.4618 

Wiedemann 
11 

25-111 

0.4280 

Mean 

0.4565 

I    O2Q 

Miiller 

0.4436 

—  28-9 

3-3996 

Regnault 

~  y 

Hydrogen  .  .  . 

12-198 

21  —  IOO 

3.4090 
3.4100 

ii 

Wiedemann 

Mean 

3  .4062 

1  .410 

Cazin 

2.419 

Nitrogen 

0-200 

0.2438 

Regnault 

1  .410 

Cazin 

0.1729 

Sulphur 
Dioxide 

16-202 

0.1544 

Regnault 

1.26 

/      Cazin      \ 
I     Miiller     / 

0.1225 

128-217 

0.4805 

Regnault 

Water  

100-125 

0-3787 

(      Gray, 
\  Macfarlane, 

Mean 

0.4296 

1.300 

Various 

0.3305 

197 


198 


APPENDIX 


TABLE  H 

DENSITY  OF  GASES 
(Taken  from  Smithsonian  Physical  Tables) 


Gas 

Specific 
gravity 

Grams  per 
cubic  centimeter 

Pounds  per 
cubic  foot 

Air        

I  .OOO 

0.001293 

0.08071 

Ammonia 

O    SQ7 

O  OOO77O 

0.04807 

Carbon  dioxide 

I    ^2O 

0.001974 

o.  12323 

Carbon  monoxide  
Coal  gas        

0.967 

o  320—0  740 

O.OOI234 
0.000414—0  000957 

0.07704 
0.02^83-0.0^973 

Hydrogen      

0.0696 

O.OOOO9O 

0.00562 

Hydrogen  sulphide.  .  . 
Marsh  gas  

I  .191 

o.zzo 

0.001476 
0.000727 

0.09214 
0.04538 

Nitrogen 

O  Q72 

O   OOI257 

o  07847 

Oxvffen 

i  io<; 

o  0014.30 

0.08027 

Sulphur  dioxide  

2  .247 

o.  002785 

0.17386 

APPENDIX 


199 


O   4j 

*s 


w 

w 


•B    -a 


111 


J»l 

w  ^ 


H    O>voO   flH 


q  q  o  o  q  o  o  o  o  o  6  5  o  o  o  o  o  o  q  o  o 
oododdddddooooooooooo 


CO  «    -^-  O^ 
O\  xo  O    t^ 


co  OO   t^O  OO    O 
O    »O  to  to  COOO    to  W    to  co  M 

co  COOO    O    COM    cotocioOtOC^ 


00 

c*    <N  r^.  CO  COOO 

t>.  M  t^.  to  ro  M 

M  M  O  O  O   O 


Tl-OOO    •*«   \f)  M    rooO    ^ 

-  1000    T}-  M    coco  O   u-j  10  t^ 

r^wooo  ^oi   M  o   O^oo 

O^  O^oo  oo  OO  oo  oo  oo  t^"  Is* 


^  10  O  co  >o\O  t>.QO  t^.M 
O  COCOM  toco^r^<N  co 
t^.M  r-».Tj-M  o^t^»o^-  ^- 
O  O  O*  O  O^OO  OO  OO  OO  t^ 


O    ^J"  O^ 
CM  ooo 


f>.  C>00  00  00  M  O  COO  <s 
CN  Tj-Q  Ov-<tf^rs-^-»OcO 
Cot^OMCoiO  >OO  t^OO 


O>MMMMM>M~<NW 

dododddoooooooooooooo 


T|-  »oO   to  co  O  •<*•  q   ^t  to  fooo   O   O  00   ^t  O  O   M   1000 

toOO     OOOMMMTflO  t^-OO  OO     O   O     O     M     M     M 

to  to  >0  toO  OOOOOOOOO   tot>.toi^t^ 


10  10  10  to  10  10  v 


rt'CN    toM    ^-toO^M    M    rfvoMO    O    COO    O*  HI 

i-^oo  oo  od<a>a*o  o  o  M  <N  «  coco<o<OTj- 


coo   O  OO   M   too  O   O   co  to  cooo  OO 


O   O 


H     M     VOO     tofOMOOO     ^H     <N     IO 

too   iOiOTfTl-Ttcococo<N    M    Q 

ooooooooooooo 


»OMOO     »OCN     O^O 

C\  OOO  00  OC    to  lo 

ooooooo 


O  M   O  tooo   c».  M   to  o 


co  M   <N   O   to  cooo   «o   O-*OO   O   tor}-OO   M   »OO 
«    «O  •*  -«t  10  iOO  OO    OO    <N    coco-^-vo  ioO  O  O 


MC<too\ 


to  co  -^-  f»    O  »0  O    rJ-00    MO    M    cow    OO    WOO    (OtoM 
co  too   to  tooO    O  O  O  O    W    TJ-  too   to  ts»oO  00    O  O  O 


M     W     CO  Tf  VOO     toOO     O 

OOOOOOOOOMWCO"*-  too   tooo    O  O    M 


200 


APPENDIX 


s.  <$  "o    • 

£>  P.  O  w 

O   O  w  O   O 

PO  to  t^OO     O  M    <N    Tt  IOO    t^OO    OO    M    M    M    CO  ^  »O  to 

•  &$$ 

oooooooooodododooodod 

II 

o  o  *>*  w  o  PO  t^oo  ^-rotot^MO  woo  ^MOOO  ^j* 

IJ 

OOOOO    OO   POM    O    OOO   t*»  t^O  O   to  to  to  rj-  rj-  ^t1 

ll 

-j- 

^  ^  to  O  O  O   M  oo   M  to  M  to  oj  oo  t^»  01   o   M  to  M   o 
\o  o  o  ^~  N  PO  oo  o  o  oo  o  POO  o  to  o  to  o  o  ^< 

O  O    to  to  PO  M    OOO    t^»vO    to  to  ^  PO  PO  01    01    M    M    O    O 

H  c 

W 

|l| 

O   PO  t^-O     IO   T}-    M     O     M     Tj-OO     O   O     Tj-VO  OO     to   M     M     t>-  t<« 

POM    ^J-MO    O    MVO    ^  ^O    O  O    PO  O  t>«O  O  O  O    *>• 
O    to  r}-  ^  OO    POOOOO    ^POM    OOO    t^O    to  rj-  PO  C» 

w° 

Pi 

05 

tOMOO     POtO<N     OOO     O     M     POO     <N     ^M     T^IOO     ^^PO 

^3 

oo-ooooooooooooooooooo 

All 

I 

NO    O  O  PO  PO  N    OO    MOO  t^-00    M   «O  t^  O    «    >0  »>• 

* 

|.i1i 

Q. 

O    POO  000000    O    M    OPOO    <NVO'^PO  toOO    PO  O    t>-O 
o*    O  t^O   toO    O  ^   to  O   to  O   to  M   t^  co  OO    PO  OO 

w    "^ 

OOO  OOOOOOOOOQOOOOOOOOOOOOOOOOOOOOOOOOOOOO 

11 

J 

O   ^  ^  t^  c»   ^j-  OOO   ^i-O  O   rh  O   tooo  OO   PO  ^t  rj-  TJ-  PO 

oo  o  o  o  o  o  POO  o  M  PO  to  t^-oo  o  M  <N  PO  TJ-  too 

tj 

•8*. 

cs    O^t^O    O    M    OPOM   toO    OO    f»   t^-POM   OOO 

IP 

^ 

rj-  H    O    O  O   <N   tooo    POOO    POO^t-M   t^POO   t^POOQO 

t^  t^  t^vO  Oto^-POPO<N«NMMMOOOOO  OOO 

OOOOOOOOOOOOOOOO  OOO  00  00  00 

"o     *o 

oo  to  O  O   M  rj-oo   o  M  ^-  M  PO  M  t^vo  Th  O   PO  to  to  PO 

ill 

4 

PO  *»•  O    M  O  OO  OO   t^-O   POOO   w   tooi   t^MvO    O   Tj-OO 
r^  t^-oo  OOOOMCNPO^to  toO  O   t^  t^OO  OO    O  O  O 

W    J 

A       P^ 

t-^  to  O 

OOioOOOMPOPOPOtoOMt^O    OO   O    PO  PO  O  00 

|5| 

O     O     M     M     CS     Tf   tO   tOO     1^*00  OOOOOOMMNMM 

Hi* 

- 

POrtTl-lOO    tOO    toO    tOO    tOO    toO    toO    IOO    toO 

MMMMCNNCOPO^^IO  *OO  O   t^  t^OO  OO   O  O  O 

APPENDIX  201 


IOCM    t^cOM    r^.  CM    t^coO    toO  COOO    COCO    CM 

CO  t^   t-^00     OOOOMCM^CMCMCOCO       " 
CM     CN     CM     CM     CM     CM 


^t  ^  to  too  l*«  t*»  t^OC  00  OOO  M  co  to 
OO  O  O  M  CM  <O  •**•  too  r^OO  O  CM  ^f  O  •"*• 
coco-<l-Tt-1<J-Tt-Tl-'<frTl--<t'<4-Tj-ioto  too 

ooododoooooooooooooooooooooood 


O    t^OO    COCM    M    OCM    CM    O    •^•COIOCM    COOO  O  O    O    **•  t^-OO    MO    ^J-  rj-  O  O    M 

CO   ^OO     CM   OO     to   CO   M     M     M     CM     CO  tO  t^    O     COO     O     "^    O   COOO     CO   O   ^"    O     CM     tOOO     to 


M    O»O    CO  M 


fOOOOOvOvOOOO    O^M    M    O^ 
^  W    O»r^xorOM    O\  r^O    ^t  O 


MOO  O  ^O  O>  w  ION  O  O^i-<O  MOO  lOTfT^-o  OfOOO  VOM  OOCO  OO*1-1 
O  O  <o  1000  O  ^-  t^  M  1000  roi^oiO  MO  MO  M  r>-<Noo  Tj-ovor^-O  M  to 
M  M  O  OOO  OO  t^O  OiO^TfrocOOtfNMMOOO  OOC  OO 
MMMOOOOOOOOOOOOOOOOO  O  O  O  O 


O  O  O  O  O  O  O 

OOOOOOOOOO 


STf  r*»  O  OOO   to  CM   i^  CM   toco    O  O  OOO  O    rf  O    *~-  fOOO    cooO    CM  O    coo    O  OO 
cot^M    toOcO»^-O    ^t^O    COO    O  CM    tOOO    M    COO  OO    M    COO  OC    co  t^OO    !*• 

•  CO   OO     O   O    O   OOMMMCMCMCMCM     CO   CO   CO    ^    ^    ^"    ^    to   to  to   toO   O     f^OO 


OOOOOOOOOOOOOOOOOOOOOOOOOOOOOO 


O  M  co  too  oo   O  O   CM  co  too  t^-oo   O  O   M  <N   co  ^  to  too  r^  t^oo  O  M  <OO 

M     CM     CM     CM     CM     CM     CM     COCOCOCOCOCOCOCO^'^''<t''^'4'^'^t''^''<t'^t'rt'vOtOlOtO 

oooooooooooooooooooooooooooooooooooooooooooooooooooooococooo 


O   t^OCMO   O   toO   t^-^-MOOO   to^-^-Tt-Tj-rf  toO  OO    O    co  tooo    ^  O    CM  OO 

CO   O     *^   to   CM     O     ^   to    CM     O   OO     to   CO   M     O  f^   to   CO   M     O   *^    to    ^    CM     O  OO     to   CM     ^O 

O    O    O  O  O  OOO  OO  OO  OO   f^»  ^  r^  t^*O  O  O  O  O   to  to  to  to  to  to  ^  ^  ^"  co  CM 

CM    OOOO    coOO   CMOO    TJ-O   toO    TfO  ^tOO   co  *^  M   toco    CM  O    O  CM    O  toco   M 

t^^OO  OO    O  O    M    M    CM    CM    co  ^  ^t"  to  to  toO  O   t^  l^OO  OO  OO    O  O  O  O    O    M    CM    ^ 

OOOOOOOO    OOOOOOOOOOOOOOOOOOOOOO    O    O    O    O 


CM   tooo    CM   r^.  co  OO    Tj-  CM    O  OC  CO   t^  t^.cO  CO    O  O    CM    rj-o    O  CM    rfoo    to  coO    co 

IOCM    or>-Tj-<N    ot^tocOMOOO    ^fcM    OOOO   IOCOM    O  r>»O    ^f  CM    OO  OO    M 
CO  CO   *^  r^  t^  t^o  OOOO   ioioiototoTj-'*4-T}-Tl-Tt-co«OcocococM    <M    M    M 

COCOOOOOOOOOOOOOOOOOOOOCOCOOOOOOOCOOOOOOOOOCOOOOOOOCOOOOOOOO 


O   to  O    co  too   t>.O    TJ-  <M    OO    «M   t>-<MO   O    Tj-t^OM    <M 

N  to  o  <N  tooo  M  Tj-  t^.  o  <M  tooo  O  co  tooo  O  «M  Tj-r>.OM  cotot^M  IO^CM 
O  O  O  M  M  M  CM  «N  <M  cocococo-^-TfTj-TttoiOtoto  too  O  O  O  t>.  1-^00  O 
------------------------  -  ------------ 


co  co  co  co 


rJ-00    M    CO^-^-COMOO    toOO    O    tooo    M    Tj-O  00    O  O    O    O    O  OOO    rj-  M    to  to 

M    rj-oo    M    -3-  I--  O    co  toOO    M    coo  00    O    co  to  f^  O  M    TfrO  00    OM    cot^-M    O»^- 
tototo  too  OOO    t>-  t»  t^  t*.  t^OO  OOOOOOOO    OOOO    O    M 


MMMMMMMMMMMC'IC^ClOldC'IC'IC'JWC^     CO 


202 


APPENDIX 


TABLE  IV.  —  PROPERTIES  OF  SUPERHEATED   STEAM 

Reproduced  by  permission  from  Marks  and  Davis'  "Steam  Tables  and  Diagrams." 
(Copyright,  1909,  by  Longmans,  Green  &  Co.) 


Pressure, 
Pounds 
Absolute. 

Satu- 
rated 
Steam. 

Degrees  of  Superheat. 

Pressure, 
Pounds 
Absolute. 

So 

100 

150 

200 

250 

300 

t 

162.3 

212.3 

262.3 

312-3 

362.3 

412.3 

462.3 

t 

5 

V 

73-3 

79-7 

85.7 

91.8 

97.8 

103.8 

109.8 

V 

5 

h 

II30-5 

II53-5 

1176.4 

II99-5 

1222.5 

1245.6 

1268.7 

h 

t 

193.2 

243.2 

293.2 

343-2 

393-2 

443-2 

493-2 

t 

10 

V 

38-4 

4i-5 

44.6 

47-7 

50.7 

53-7 

56.7 

V 

10 

h 

1143.1 

1166.3 

1189.5 

1212  .7 

1236.0 

1259-3 

1282.5 

h 

t 

213  .0 

263  .0 

313.0 

363-0 

4i3-o 

463-0 

513-0 

t 

15 

V 

26.27 

28.40 

30.46 

32.50 

34-53 

36.56 

38.58 

V 

15 

h 

1150.7 

1174.2 

1197.6 

1221  .O 

1244.4 

1267.7 

1291  .1 

h 

t 

228.0 

278.0 

328.0 

378.0 

428.0 

478.0 

528.0 

t 

20 

V 

20.08 

21  .69 

23.25 

24.80 

26.33 

27-85 

29-37 

V 

20 

h 

1156.2 

II79.9 

1203.5 

I227.I 

1250.6 

1274.1 

1297.6 

h 

t 

240.1 

29O.I 

340.1 

390.1 

440.1 

490.1 

540.1 

t 

25 

V 

16.30 

17.60 

18.86 

20.10 

21.32 

22.55 

23-77 

V 

25 

h 

1160.4 

1184.4 

I  2O8  .  2 

1231.9 

1255-6 

1279.2 

1302.8 

h 

t 

250.4 

300.4 

350.4 

400.4 

450.4 

500.4 

550.4 

t 

3° 

v' 

13-74 

14.83 

15.89 

16.93 

17.97 

18.99 

20.00 

V 

30 

h 

1163.9 

II88.I 

I2I2.I 

1236.0 

1259-7 

1283.4 

1307.1 

h 

t 

259-3 

309.3 

359-3 

409.3 

-459-3 

509-3 

559-3 

t 

35 

V 

11.89 

12.85 

13-75 

14.65 

15-54 

16.42 

17.30 

V 

35 

h 

1166.8 

II9I.3 

1215.4 

1239.4 

1263.3 

1287.1 

1310.8 

h 

t 

267.3 

317.3 

367.2 

4I7.3 

467-3 

517-3 

567-3 

t 

40 

V 

10.49 

"•33 

12.13 

12.93 

13.70 

14.48 

15-25 

V 

40 

h 

1169.4 

1194.0 

1218.4 

1242.4 

1266.4 

1290.3 

1314.1 

h 

t 

274-5 

324-5 

374-5 

424.5 

474-5 

524-5 

574-5 

t 

45 

V 

9-39 

10.14 

10.86 

n-57 

12.27 

12  .96 

13-65 

V 

45 

h 

1171  .6 

1196.6 

1221  .O 

1245.2 

1269.3 

1293-2 

1317.0 

h 

t 

281.0 

331-0 

381.0 

431.0 

481.0 

531-0 

581.0 

t 

50 

V 

8.51 

9.19 

9.84 

10.48 

ii  .11 

11.74 

12.36 

V 

50 

h 

1173.6 

1198.8 

1223.4 

1247.7 

1271.8 

1295.8 

I3I9-7 

h 

t 

287.1 

337-1 

387.1 

437-1 

487.1 

537-1 

587-1 

t 

55 

V 

7-78 

8.40 

9.OO 

9-59 

10.  16 

10.73 

11.30 

V 

55 

h 

H75-4 

1200.8 

1225.6 

1250.0 

1274.2 

1298.1 

1322.0 

h 

t 

292.7 

342-7 

392.7 

442.7 

492.7 

542.7 

592.7 

t 

60 

V 

7.17 

7-75 

8.30 

8.84 

9-36 

9.89 

10.41 

V 

60 

h 

1177.0 

i  202  .6 

1227.6 

1252.1 

1276.4 

1300.4 

1324-3 

h 

t 

298.0 

348.0 

398.0 

448.0 

498.0 

548.0 

598.0 

t 

65 

V 

6.65 

7.20 

7.70 

8.20 

8.69 

9.17 

9-65 

V 

65 

h 

1178.5 

1204.4 

1229.5 

1254.0 

1278.4 

1302.4 

1326.4 

h 

t 

302.9 

352.9 

402.9 

452.9 

502.9 

552.9 

602  .9 

t 

70 

V 

6.  20 

6.71 

7-18 

7.65 

8.  ii 

8.56 

9.01 

V 

'  70 

h 

1179.8 

1205.9 

I23I.2 

1255.8 

1280.2 

1304-3 

1328.3 

h 

t 

307.6 

357-6 

407.6 

457-6 

507.6 

557-6 

607.6 

t 

75 

V 

5-8i 

6.28 

6-73 

7.17 

7.60 

8.02 

8-44 

V 

75 

h 

1181.1 

1207.5 

1232.8 

1257-5 

1282.0 

1306.1 

1330.1 

h 

t 

312.0 

362.0 

412  .0 

462  .0 

512.0 

562.0 

612  .0 

t 

80 

V 

5-47 

5-92 

6-34 

6-75 

7.17 

7.56 

7-95 

V 

80 

h 

1182.3 

1208.8 

1234-3 

1259.0 

1283.6 

1307.8 

I33L9 

h 

t 

3*6.  3 

366.3 

416.3 

466.3 

516.3 

566.3 

616.3 

t 

85 

V 

5-i6 

5-59 

6-99 

6.38 

6.76 

7-14 

7-51 

V 

85 

h 

1183.4 

I2I0.2 

1235-8 

1260.6 

1285.2 

1309.4 

1333-5 

h 

Temperature,  deg.  Fahr.  v  =  Specific  volume,  in  cubic  feet,  per  pound. 

h  =  Total  heat  from  water  at  32  degrees,  B.t.u. 


APPENDIX 

TABLE  IV.  —  Continued 


203 


Pressure, 
Pounds 
Absolute. 

Satu- 
rated 
Steam. 

Degrees  of  Superheat. 

Pressure, 
Pounds 
Absolute. 

So 

IOO 

150 

200 

250 

300 

t 

320-3 

370-3 

420.3 

470-3 

520.3 

570.3 

620.3 

/ 

QO 

V 

4.89 

5-29 

5.67 

6.04 

6.4O 

6.76 

7.11 

V 

go 

h 

1184.4 

I2II  .4 

1237.2 

1262  .0 

1286.6 

1310.8 

J334-9 

h 

t 

324-1 

374-1 

424.1 

474-1 

S24.I 

574-1 

624.1 

t 

95 

V 

4-65 

5-03 

5-39 

5-74 

6.O9 

6-43 

6.76 

V 

95 

h 

1185.4 

I2I2.6 

1238.4 

1263.4 

I288.I 

1312.3 

1336.4 

h 

t 

327.8 

377-8 

427.8 

477-8 

527.8 

577-8 

627.8 

t 

IOO 

V 

4-43 

4-79 

5-14 

5-47 

5-80 

6.12 

6-44 

V 

IOO 

h 

1186.3 

1213.8 

1239.7 

1264.7 

1289.4 

1313-6 

1337-8 

h 

t 

331-4 

381.4 

431-4 

481.4 

531-4 

581.4 

631-4 

t 

105 

V 

4-23 

4.58 

4.91 

5-23 

5-54 

5-85 

6-15 

V 

105 

h 

1187.2 

1214.9 

1240.8 

1265.9 

i  290  .  6 

1314-9 

I339-I 

h 

t 

334-8 

384-8 

434-8 

484.8 

534-8 

584-8 

634-8 

t 

no 

V 

4-05 

4-38 

4.70 

5  -oi 

5-31 

5-6i 

5-90 

V 

no 

h 

1188.0 

1215.9 

1242.0 

1267.1 

1291.9 

1316.2 

1340-4 

h 

t 

338.1 

388.1 

438.1 

488.1 

538.1 

588.1 

638-1 

t 

115 

V 

3-88 

4.20 

4-51 

4.81 

5-09 

5.38 

5-66 

i' 

"5 

h 

1188.8 

1216.9 

1243.1 

1268.2 

1293.0 

1317-3 

I34L5 

h 

t 

341-3 

391-3 

441-3 

491-3 

541-3 

591-3 

641-3 

t 

120 

V 

3-73 

4.04 

4-33 

4.62 

4.89 

5-17 

5-44 

v 

120 

h 

1189.6 

1217.9 

1244.1 

1269.3 

1294.1 

1318.4 

1342.7 

h 

t 

344-4 

394-4 

444-4 

494-4 

544-4 

594-4 

644.4 

t 

125 

V 

3.58 

3-88 

4-17 

4-45 

4-71 

4-97 

5-23 

v 

125 

h 

1190.3 

1218.8 

1245.1 

1270.4 

1295.2 

1319-5 

1343-8 

h 

t 

347-4 

397-4 

447-4 

497-4 

547-4 

597-4 

647-4 

t 

130 

V 

3-45 

3-74 

4.02 

4.28 

4-54 

4.80 

5-05 

v 

130 

h 

1191  .0 

1219.7 

1246.1 

1271.4 

1296.2 

1320.6 

1344-9 

h 

t 

350.3 

400.3 

450-3 

500.3 

550.3 

600.3 

650.3 

t 

135 

v 

3-33 

3-6i 

3-88 

4-14 

4.38 

4-63 

4.87 

v 

J35 

h 

1191  .6 

1220.6 

1247.0 

1272.3 

1297.2 

1321.6 

1345-9 

h 

• 

t 

353-1 

403.I 

453-1 

503-1 

553-1 

603.1 

653-1 

t 

140 

V 

3-22 

3-49 

3-75 

4.00 

4-24 

4.48 

4-71 

V 

140 

h 

1192  .2 

1221.4 

1248.0 

1273-3 

I  298  .  2 

1322.6 

1346.9 

h 

t 

355-8 

405.8 

455-8 

505-8 

555-8 

605.8 

655.8 

t 

145 

V 

3.12 

3-38 

3-63 

3-87 

4.10 

4-33 

4.56 

v 

J45 

h 

1192.8 

1222.2 

1248.8 

1274.2 

1299.1 

1323.6 

1347-9 

h 

t 

358.5 

408.5 

458.5 

508.5 

558.5 

608.5 

658-5 

t 

150 

V 

3.01 

3-27 

3-5i 

3-75 

3-97 

4.19 

4.41 

V 

J5o 

h 

IIQ3-4 

1223.0 

i  249  .  6 

1275.1 

1300.0 

1324-5 

1348".  8 

h 

t 

361.0 

411  .O 

461  .0 

511  .0 

561  .0 

611  .0 

661.0 

t 

155 

V 

2.92 

3-17 

3-4i 

3-63 

3-85 

4.06 

4.28 

V 

i55 

h 

1194.0 

1223.6 

1250-5 

1276.0 

1300.8 

1325-3 

1349-7 

h 

t 

363-6 

413.6 

463.6 

5I3-6 

563-6 

613.6 

663.6 

t 

1  60 

V 

2.83 

3-07 

3-3o 

3-53 

3-74 

3-95 

4-i5 

v 

1  60 

h 

II94-5 

1224.5 

1251-3 

1276.8 

1301.7 

1326.2 

1350-6 

h 

t 

366.0 

416.0 

466.0 

516.0 

566.0 

616.0 

666.0 

t 

165 

V 

2-75 

2.99 

3-21 

3-43 

3-64 

3-84 

4-04 

v 

165 

h 

1195.0 

1225.2 

1252.0 

1277.6 

1302-5 

1327.1 

I35I-5 

h 

t 

368.5 

418.5 

468.5 

518.5 

568.5 

618.5 

668.5 

: 

170 

V 

2.68 

2.9I 

3.12 

3-34 

3-54 

3-73 

3-92 

v 

170 

h 

"95-4 

1225.9 

1252.8 

1278.4 

I303-3 

1327.9 

1352.3 

h 

t  =  Temperature,  deg.  Fahr.  v  =  Specific  volume,  in  cubic  feet,  per  pound. 

h  =  Total  heat  from  water  at  32  degrees,  B.t.u. 


204 


APPENDIX 

TABLE  IV.  —  Continued 


Pressure, 
Pounds 
Absolute. 

Satu- 
rated 
Steam. 

Degrees  of  Superheat. 

Pressure, 
Pounds 
Absolute 

50 

100 

150 

200 

250 

300 

t 

370.8 

420.8 

470.8 

520.8 

570.8 

620.8 

670.8 

t 

175 

V 

2.60 

2-83 

3-04 

3-24 

3-44 

3-63 

3.82 

V 

175 

h 

HQ5-9 

1226.6 

1253-6 

I279.I 

1304.1 

1328.7 

1353-2 

h 

t 

373-1 

423-I 

473-1 

523  -1 

573-1 

623.1 

673.1 

t 

180 

V 

2-53 

2-75 

2  .96 

3-16 

3-35 

3-54 

3.72 

V 

1  80 

h 

1196.4 

1227.2 

1254.3 

1279.9 

1304.8 

1329-5 

1353.9 

h 

t 

375-4 

425-4 

475-4 

525-4 

575-4 

625.4 

675-4 

t 

185 

V 

2.47 

2.68 

2.89 

3.08 

3-27 

3-45 

3.63 

V 

185 

h 

1196.8 

1227.9 

1255-0 

1280.6 

1305-6 

1330.2 

1354.7 

h 

t 

377-6 

427.6 

477.6 

527.6 

577-6 

627.6 

677.6 

t 

190 

V 

2.41 

2.62 

2.81 

3-00 

3-i9 

3-37 

3-55 

V 

190 

h 

H97-3 

1228.6 

1255-7 

1281.3 

1306.3 

1330.9 

1355-5 

h 

t 

379-8 

429.8 

479-8 

529-8 

579-8 

629.8 

679.8 

t 

iQS 

V 

2-35 

2-55 

2-75 

2-93 

3-ii 

3-29 

3-46 

V 

i95 

h 

1197.7 

1229.2 

1256.4 

1282.0 

1307.0 

I33I-6 

1356.2 

h 

t 

381.9 

43i-9 

481.9 

531-9 

581.9 

631.9 

681.9 

t 

200 

V 

2.29 

2.49 

2.68 

2.86 

3-04 

3-21 

3.38 

V 

200 

h 

1198.1 

1229.8 

1257.1 

1282.6 

1307.7 

1332-4 

1357-0 

h 

t 

384.0 

434-0 

484.0 

534-o 

584.0 

634.0 

684.0 

t 

205 

V 

2.24 

2.44 

2.62 

2.80 

2.97 

3-i4 

3-30 

V 

205 

h 

1198.5 

1230.4 

1257-7 

1283.3 

1308.3 

i333-o 

1357-7 

h 

t 

386.0 

436.0 

486.0 

536.0 

586.0 

636.0 

686.0 

t 

210 

V 

2.19 

2.38 

2-56 

2.74 

2.91 

3-07 

3-23 

V 

2IO 

h 

1198.8 

1231.0 

1258.4 

1284.0 

1309.0 

1333-7 

1358.4 

h 

t 

388.0 

438.0 

488.0 

538.o 

588.0 

638.0 

688.0 

t 

215 

V 

2.14 

2-33 

2-51 

2.68 

2.84 

3.00 

3-i6 

V 

215 

-h 

1199.2 

1231.6 

1259.0 

1284.6 

1309.7 

1334-4 

I359-I 

h 

t 

389-9 

439-9 

489.9 

539-9 

589-9 

639-9 

689.9 

t 

220 

V 

2  .09 

2.28 

2-45 

2.62 

2.78 

2-94 

3.10 

V 

22O 

h 

II99.6 

1232.2 

1259.6 

1285.2 

1310.3 

!335-i 

1359.8 

h 

t 

39J-9 

441.9 

491.9 

541-9 

591-9 

641.9 

691.9 

t 

225 

V 

2.05 

2.23 

2  .40 

2-57 

2.72 

2.88 

3-03 

V 

225 

h 

1199.9 

1232.7 

1260.2 

1285.9 

1310.9 

1335-7 

1360.3 

h 

t 

393-8 

443-8 

493-8 

543-8 

593-8 

643-8 

693.8 

t 

230 

V 

2  .00 

2.18 

2-35 

2-51 

2.67 

2.82 

2.97 

V 

230 

h 

1200.2 

1233.2 

1260.7 

1286.5 

1311  .6 

1336.3 

1361  .0 

h 

t 

395-6 

445-6 

495-6 

545-6 

595-6 

645-6 

695-6 

t 

235 

V 

1.96 

2.14 

2.30 

2.46 

2.62 

2-77 

2.91 

V 

235 

h 

I  200  .  6 

1233-8 

1261  .4 

1287.1 

1312.2 

1337-0 

1361.7 

h 

t 

397-4 

447-4 

497-4 

547-4 

597-4 

647.4 

697.4 

t 

240 

V 

1.92 

2.09 

2.26 

2.42 

2-57 

2.71 

2.85 

V 

24O 

h 

I  200  .  9 

1234-3 

1261  .9 

1287.6 

1312.8 

1337-6 

1362.3 

h 

t 

399-3 

449-3 

499-3 

549-3 

599-3 

649-3 

699-3 

t 

245 

V 

1.89 

2.05 

2  .22 

2-37 

2.52 

2.66 

2.80 

V 

245 

h 

I2OI  .2 

1234.8 

1262.5 

1288.2 

i3I3-3 

1338.2 

1362.9 

h 

t 

4OI  .O 

451-0 

501  .0 

55i-o 

601.0 

651  .0 

701  .0 

t 

250 

V 

1-85 

2  .02 

2.17 

2-33 

2-47 

2.61 

2-75 

V 

250 

h 

I20I.5 

1235.4 

1263.0 

1288.8 

I3I3-9 

1338.8 

1363-5 

h 

t 

402.8 

452.8 

502.8 

552-8 

602.8 

652.8 

702.8 

t 

255 

V 

1.81 

1.98 

2.14 

2.28 

2-43 

2.56 

2.70 

V 

255 

h 

I2OI  .8 

1235-9 

1263.6 

1289.3 

I3I4-5 

1339-3 

1364.1 

h 

Temperature,  deg.  Pahr.  v  =  Specific  volume,  in  cubic  feet,  per  pound. 

h  =  Total  heat  from  water  at  32  degrees,  B.t.u. 


APPENDIX 


205 


TABLE  V.  —  SATURATED  VAPOR  OF  SULPHUR  DIOXIDE 

Reproduced  by  permission  from  Peabody's  "  Steam  and  Entropy  Tables." 
English  Units. 


Temperature, 
Decrees  Fah- 
renheit. 

Pressure, 
Pounds  per 
Square  Inch. 

Heat  of  the 
Liquid. 

1 
I 

Heat  of 
Vaporization. 

Heat  Equivalent 
of  Internal 
Work. 

Heat  Equivalent 
of  External 
Work. 

Entropy  of  the 
Liquid. 

Specific  Volume. 

t 

P 

h 

H 

L 

P 

Apu 

e 

5 

-40 

3-14 

-29 

1  66 

195 

182 

13 

—  0.0632 

23.0 

—  2O 

5-90 

—  21 

169 

190 

I76 

14 

-0.0447 

12.7 

o 

10-35 

-13 

172 

i85 

170 

15 

-0.0268 

7-54 

IO 

I3-4I 

9 

i73 

182 

167 

15 

—  0.0182 

5-93 

20 

17-15 

-  5 

174 

179 

164 

15 

—  0.0098 

4.72 

30 

21.66 

—  i 

176 

177 

162 

15 

—  0.0016 

3-8i 

40 

27.06 

3 

177 

174 

158 

16 

0.0064 

3.10 

50 

33-45 

7 

178 

171 

155 

16 

0.0144 

2.58 

60 

40.98 

ii 

179 

1  68 

152 

16 

0.0221 

2.  II 

70 

49-75 

15 

181 

166 

150 

16 

0.0297 

I.78 

TABLE  VI.  —  PROPERTIES  OF  CARBON  DIOXIDE 

Reproduced  from  Marks'  "  Mechanical  Engineers'  Handbook." 


>  • 

c 

o> 

rt    r/»"£ 

.6-8 

§«£ 

Heat  Content 

o| 

|| 

11 

'3jHj 

g 
a 

•3 

"o  . 
|-3 

^     3J     C 

W3   C   ^ 

§5  i^ 

"a  o 

r^  *  ^ 

Q 

S  cr 

l!a 

III 

of 
Liquid. 

of 
Vapor. 

"1 

c  c. 

^ 

1 

I3 

/ 

£ 

h 

H 

L 

P 

Apu 

S 

9 

20 

221.0 

—  24  •  75 

100.50 

125   25 

109.0 

16.3 

0.4173 

-0.0513 

0 

308.0 

—  16.00 

IOI  .OO 

117.00 

101.3 

15.7 

O.29l8 

-0.0325 

10 

362.5 

—  11.36 

100.89 

112.35 

96.9 

15-4 

o  .  2450 

—  O.O227 

20 

421.6 

—  6.40 

100.43 

106.83 

91.8 

15-0 

0.2060 

—  0.0126 

30 

488.8 

-   1.04 

99-43 

100.47 

86.7 

14.2 

0.1724 

—  O.OO2I 

40 

564-5 

4-36 

98.25 

93-89 

80.4 

13-5 

0.1444 

O.OO87 

50 

650.0 

10.76 

96.30 

85-54 

73-31 

12.2 

o  .  i  205 

0.0205 

60 

744.0 

17-85 

93-54 

75-69 

64.90 

10.8 

0.0986 

0-0334 

70 

847.0 

26.02 

89.35 

63-33 

54-03 

9-3 

0.0816 

0.0483 

TABLE  VII.  —  NAPERIAN  LOGARITHMS 

Reproduced  by  permission  from  Goodenough's  "  Properties  of  Steam  and  Ammonia." 

e  =2.7182818  log  e  =  0.4342945  =•  M 


0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1.0 

0.  0000 

0.00995 

0.0198 

0.0295 

0.  0392 

0.0487£ 

0.0582 

0.  0676 

0.0769 

0.08618 

1.1 

1.2 
1.3 

0.0953 
0.  1823 
0.2624 

0.  1044 
0.  1906 
0.  2700 

0.1133 
0.  1988 
0.2776 

0.  1222 
0.2070 
0.2852 

0.1310 
0.2151 
0.2927 

0.1398 
0.2231 
0.3001 

0.  1484 
0.2311 
0.3075 

0.  1570 
0.2390 
0.3148 

0.  1655 
0.  2469 
0.3221 

0.1739 
0.2546 
0.3293 

1.4 
1.5 
1.6 

0.  3365 
0.  4055 
0.4700 

0.3436 
0.4121 
0.4762 

0.3507 
0.4187 
0.4824 

0.3577 
0.4253 
0.4886 

0.3646 
0.4318 
0.4947 

0.3716 
0.4382 
Q.  5008 

0.3784 
0.  4447 
0.5068 

0.3853 
0.4511 
0.5128 

0.  3920 
0.4574 
0.5188 

0.3988 
0.4637 
0.  5247 

1.7 
1.8 
1.9 

0.5306 
0.5878 
0.6418 

0.5365 
0.5933 
0.6471 

0.5423 
0.  5988 
0.6523 

0.5481 
0.  6043 
0.6575 

0.5539 
0.6098 
0.6627 

0.5596 
0.6152 
0.6678 

0.  5653 
0.6206 
0.6729 

0.5710 
0.6259 
0.  6780 

0.  5766 
0.6313 
0.6831 

0.5822 
0.6366 
0.6881 

2.0 

0.6931 

0.6981 

0.7031 

0.  7080 

0.7129 

0.7178 

0.7227 

0.7275 

0.7324 

0.7372 

2.1 
2.2 
2.3 

0.7419 
0.  7884 
0.8329 

0.7467 
0.7930 
0.8372 

0.7514 
0.7975 
0.8416 

0.7561 
0.  8020 
0.8459 

0.7608 
0.8065 
0.  8502 

0.7655 
0.8109 
0.8544 

0.7701 
0.8154 
0.  8587 

0.7747 
0.8198 
0.  8629 

0.7793 
0.  8242 
0.8671 

0.  7839 
0.8286 
0.8713 

2.4 
2.5 
2.6 

0.8755 
0.9163 
0.9555 

0.8796 
0.9203 
0.9594 

0.8838 
0.9243 
0.  9632 

0.8879 
0.9282 
0.9670 

0.  8920 

0.9322 
0.9708 

0.8961 
0.9361 
0.9746 

0.9002 
0.9400 
0.9783 

0.9042 
0.9439 
0.9821 

0.  9083 
0.9478 
0.9858 

0.9123 
0.9517 
0.9895 

2.7 
2.8 
2.9 

0.9933 
.0296 
.0647 

0.9969 
1.0332 
1.0682 

1.0006 
1.0367 
1.0716 

1.0043 
1.0403 
1.0750 

1.0080 
1.0438 
1.0784 

1.0116 
1.0473 
1.0818 

1.0152 
1.0508 
1.0852 

1.0188 
1.0543 
1.0886 

1.0225 
1.0578 
1.0919 

1.0260 
1.0613 
1.0953 

3.0 

.0986 

1.1019 

.1053 

1.1086 

.1119 

1.1151 

1.1184 

1.1217 

1.1249 

1.1282 

3.1 
3.2 
3.3 

.1314 
.1632 
.1939 

1.1346 
1.1663 
1.1969 

.1378 
1.1694 
1.2000 

1.1410 
1.1725 
1.2030 

.1442 
.1756 
.2060 

1.1474 
1.1787 
.2090 

1.1506 
1.1817 
1.2119 

1.1537 
1.1848 
1.2149 

1.1569 
1.1878 
1.2179 

1.1600 
1.1909 
1.2208 

3.4 
3.5 
3.6 

.2238 
.2528 
.2809 

.2267 
.2556 
.2837 

1.2296 
1.2585 
1.2865 

.2326 
.2613 
.2892 

.2355 
.2641 
.2920 

.2384 
.2669 
.2947 

1.2413 
1.2698 
1.2975 

1.2442 
1.2726 
1.3002 

1.2470 

1.2754 
1.3029 

1.2499 
1.2782 
1.3056 

3.7 
3.8 
3.9 

.3083 
.3350 
.3610 

.3110 
.3376 
.3635 

1.3137 
1.3403 
1.3661 

.3164 
.3429 
.3686 

.3191 
.3455 
.3712 

.3218 
.3481 
.3737 

1.3244 
1.3507 
1.3762 

1.3271 
1.3533 
1.3788 

1.3297 
1.3558 
1.3813 

1.3324 
1.3584 
1.3838 

4.0 

.3863 

.3888 

1.3913 

.3938 

.3962 

.3987 

1.4012 

.4036 

1.4061 

.4085 

4.1 
4.2 
4.3 

.4110 
.4351 
.4586 

.4134 
.4375 
.4609 

1.4159 
1.4398 
1.4633 

.4183 
.4422 
.4656 

.4207 
.4446 
.4679 

.4231 
.4469 
.4702 

1.4255 
1.4493 
1.4725 

.4279 
.4516 

.4748 

1.4303 
1.4540 
1.4770 

.4327 
.4563 
.4793 

4.4 
4.5 
4.6 

.4816 
.5041 
.5261 

.  4839 
.5063 
.5282 

1.4861 
1.5085 
1.5304 

.4884 
.5107 
.5326 

.4907 
.5129 
.5347 

.4929 
.5151 
.5369 

1.4951 
1.5173 
1.5390 

.4974 
.5195 
.5412 

.4996 
.5217 
.  5433 

.5019 
.5239 
.5454 

4.7 
4.8 
4.9 

.5476 
.5686 
.5892 

.5497 
.5707 
.5913 

1.5518 
1.5728 
1.5933 

.5539 
.5748 
.5953 

.5560 
.5769 
.5974 

.5581 
.5790 
.5994 

1.5602 
1.5810 
1.6014 

.5623 
.5831 
.6034 

.5644 
.5851 
.6054 

.5665 

.5872 
.6074 

5.0 

.6094 

.6114 

1.6134 

.6154 

.6174 

.6194 

1.6214 

.6233 

.6253 

.6273 

6.1 
5.2 
5.3 

.6292 
.6487 
.6677 

.6312 
.6506 
.6696 

1.6332 
1.6525 
1.6715 

.6351 
.6544 
.6734 

.6371 
.6563 
.6752 

.6390 
.6582 
.6771 

1.6409 
1.6601 
1.6790 

.6429 
.6620 
.6808 

.6448 
.6639 
.6827 

.6467 
.6658 
.6845 

5.4 
5.5 
5.6 

.6864 
.7047 
.7228 

.6882 
.7066 
.7246 

1.6901 
1.7884 
1.7263 

.6919 
.7102 
.7281 

.6938 
.7120 
.7299 

.6956 
.7138 
.7317 

1.6974 
1.7156 
1.7334 

.6993 
.7174 
.7352 

.7011 
.7192 
.7370 

.7029 
.7210 
.7387 

206 


TABLE  VII.  —  Continued.    NAPERIAN  LOGARITHMS 


0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

6.7 
6.0 
5.9 

1.7405 
1.7579 
1.7750 

1.7422 
1.7596 
1.7766 

1.7440 
1.7613 
1.7783 

1.7457 
1.7630 
1.7800 

1.7475 
1.7647 
1.7817 

1.7492 
1.7664 
1.7834 

1.7509 
1.7681 
1.7851 

1.7527 
1.7699 
1.7867 

1.7544 
1.7716 
1.7884 

1.7561 
1.7733 
'1.7901 

6.0 

1.7918 

1.7934 

1.7951 

1.7967 

1.7984 

1.8001 

1.8017 

1.8034 

1.8050 

1.8066 

6.1 
6.2 
6.3 

1.8083 
1.8245 
1.8405 

1.8099 
1.8262 
1.8421 

.8116 

.8278 
.8437 

1.8132 
1.8294 
1.8453 

1.8148 
1.8310 
1.8469 

1.8165 
1.8326 
1.8485 

1.8181 
1.8342 
1.8500 

1.8197 
1.8358 
1.8516 

1.8213 
1.8374 
1.8532 

1.8229 
1.8390 
1.8547 

6.4 
6.5 
6.6 

1.8563 
1.8718 
1.8871 

1.8579 
1.8733 
1.8886 

.8594 
.8749 
1.8901 

1.8610 
1.8764 
1.8916 

1.8625 
1.8779 
1.8931 

1.8641 
1.8795 
1.8946 

1.8656 
1.8810 
1.8961 

1.8672 
1.8825 
1.8976 

1.8687 
1.8840 
1.8991 

1.8703 
1.8856 
1.9006 

6.7 
6.8 
6.9 

1.9021 
1.9169 
1.9315 

1.9036 
1.9184 
1.9330 

1.9051 
1.9199 
1.9344 

1.9066 
1.9213 
1.9359 

1.9081 
1.9228 
1.9373 

1.9095 
1.9242 
1.9387 

1.9110 
1.9257 
1.9402 

1.9125 
1.9272 
1.9416 

1.9140 
1.9286 
1.9430 

1.9155 
1.9301 
1.9445 

7.0 

1.9459 

1.9473 

1.9488 

1.9502 

1.9516 

1.9530 

1.9544 

1.9559 

1.9573 

1.9587 

7.1 
7.2 
7.3 

1.9601 
1.9741 
1.9879 

1.9615 
1.9755 
1.9892 

1.9629 
1.9769 
1.9906 

1.9643 

1.9782 
1.9920 

1.9657 
1.9796 
1.9933 

1.9671 
1.9810 
1.9947 

1.9685 
1.9824 
1.9961 

1.9699 
1.9838 
1.9974 

1.9713 
1.9851 
1.9988 

1.9727 
1.9865 
2.  0001 

7.4 
7.5 
7.6 

2.0015 
2.0149 
2.  0281 

2.  0028 
2.0162 
2.  0295 

2.0042 
2.0176 
2.0308 

2.  0055 
2.0189 
2.0321 

2.0069 
2.0202 
2.0334 

2.0082 
2.0215 
2.0347 

2.0096 
2.0229 
2.0360 

2.0109 
2.0242 
2.  0373 

2.0122 
2.0255 
2.0386 

2.0136 
2.0268 
2.0399 

7.7 
7.8 
7.9 

2.0412 
2.0541 
2.0668 

2.0425 
2.0554 
2.0681 

2.  0438 
2.0567 
2.0694 

2.0451 
2.0580 
2.0707 

2.  0464 
2.0592 
2.0719 

2.0477 
2.0605 
2.0732 

2.  0490 
2.0618 
2.0744 

2.0503 
2.0631 
2.  0757 

2.0516 
2.0643 
2.0769 

2.0528 
2.0656 
2.  0782 

8.0 

2.  0794 

2.  0807 

2.0819 

2.0832 

2.0844 

2.0857 

2.0869 

2.0881 

2.0894 

2.  0906 

8.1 
8.2 
8.3 

2.0919 
2.1041 
2  1163 

2.0931 
2.1054 
2.1175 

.0943 
.1066 
.1187 

2.0956 
2.1078 
2.1199 

2.  0968 
2.  1090 
2.1211 

2.0980 
2.1102 
2.1223 

2.  0992 
2.1114 
2.  1235 

2.  1005 
2.1126 
2.  1247 

2.1017 
2.1138 
2.  1258 

2.  1029 
2.1150 
2.1270 

8.4 
8.5 
8.6 

2.  1282 
2.1401 
2.1518 

2.  1294 
2.1412 
2.  1529 

.1306 
.1424 
.1541 

2.1318 
2.1436 
2.  1552 

2.  1330 
2.  1448 
2.  1564 

2.  1342 
2.  1459 
2.1576 

2.  1353 
2.  1471 
2.  1587 

2.1365 
2.  1483 
2.  1599 

2.  1377 
2.  1494 
2.1610 

2.  1389 
2.  1506 
2.  1622 

8.7 

8.8 
8.9 

2.1633 
2.  1748 
2.1861 

2.  1645 
2.1759 
2.1872 

.1656 
.1770 
.1883 

2.1668 
2.  1782 
2.  1894 

2.  1679 
2.1793 
2.  1905 

2.  169*1 
2.1804 
2.1917 

2.1702 
2.1815 
2.1928 

2.1713 
2.1827 
2.  1939 

2.1725 
2.  1838 
2.  1950 

2.  1736 
2.  1849 
2.1961 

9.0 

2.  1972 

2.  1983 

.1994 

2.  2006 

2.2017 

2.2028 

2.2039 

2.2050 

2.2061 

2.2072 

9.1 
9.2 
9.3 

2.  2083 
2.2192 
2.2300 

2.  2094 
2.2203 
2.2311 

.2105 
.2214 
2.  2322 

2.2116 
2.2225 
2.2332 

2.2127 
2.2235 
2.2343 

2.2138 
2.2246 
2.2354 

2.2148 
2.2257 
2.2364 

2.2159 
2.2268 
2.2375 

2.2170 
2.2279 
2.2386 

2.2181 
2.  2289 
2.2396 

9.4 
9.5 
9.6 

2.  2407 
2.2513 
2.2618 

2.2418 
2.  2523 
2.2628 

2.  2428 
2.2534 
2.2638 

2.  2439 
2.2544 
2.2649 

2.  2450 
2.  2555 
2.2659 

2.  2460 
2.2565 
2.2670 

2.2471 
2.2576 
2.  2680 

2.  2481 
2.  2586 
2.  2690 

2.  2492 
2.2597 
2.2701 

2.2502 
2.  2607 
2.2711 

9.7 

9.8 
9.9 

2.2721 

2.2824 
2.2925 

2.  2732 
2.2834 
2.2935 

2.  2742 
2.2844 
2.2946 

2.  2752 
2.  2854 
2.  2956 

2.  2762 
2.2865 
2.2966 

2.2773 
2.  2875 
2.2976 

2.  2783 
2.  2885 
2.2986 

2.  2793 

2.2895 
2.  2996 

2.2803 
2.  2905 
2.3006 

2.2814 
2.2915 
2.3016 

10.0 

2.3026 

207 


TABLE  VIII.  —  LOGARITHMS 

Reproduced  by  permission  from  Goodenough's  "Properties  of  Steam  and  Ammonia.' 


Proportional  Parts. 

Nat. 
Nos. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

123 

456 

789 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

4  8  12 

17  21  25 

29  33  37 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

4  8  11 

15  19  23 

26  30  34 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

3  7  10 

14  17  21 

24  28  31 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

3  6  10 

13  16  19 

23  26  29 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

369 

12  15  18 

21  24  27 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

368 

11  14  17 

20  22  25 

16 

2041 

2068 

2095 

2J22 

2148 

2175 

2201 

2227 

2253 

2279 

358 

11  13  16 

18  21  24 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

257 

10  12  15 

17  20  22 

18 

2553 

2577 

2601 

%25 

2648 

2672 

2695 

2718 

2742 

2765 

257 

9  12  14 

16  19  21 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

247 

9  11  13 

16  18  20 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

246 

8  11  13 

15  17  19 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

246 

8  10  12 

14  16  18 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

246 

8  10  12 

14  15  17 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

246 

7  9  11 

13  15  17 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

245 

7  9  11 

12  14  16 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

235 

7  9  10 

12  14  15 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

235 

7  8  10 

11  13  15 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

235 

689 

11  13  14 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

235 

689 

11  12  14 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

1  3  4 

679 

10  12  13 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

1  3  4 

679 

10  11  13 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

1  3  4 

678 

10  11  12 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

1  3  4 

578 

9  11  12 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

1  3  4 

568 

9  10  12 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

1  3  4 

568 

9  10  11 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

1  2  4 

567 

9  10  11 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

1  2  4 

567 

8  10  11 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

1  2  3 

567 

8  9  10 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

1  2  3 

567 

8  9  10 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

1  2  3 

457 

8  9  10 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

1  2  3 

456 

8  9  10 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

1  2  3 

456 

789 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

1  2  3 

456 

789 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

1  2  3 

456 

789 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

1  2  3 

456 

789 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

1  2  3 

456 

789 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

1  2  3 

456 

778 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

1  2  3 

455 

678 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

1  2  3 

445 

678 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

1  2  3 

445 

678 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

1  2  3 

345 

678 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

1  2  3 

345 

678 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

1  2  2 

345 

6  7  7 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

1  2  2 

345 

667 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

122 

345 

667 

208 


TABLE  VIII.  —  Continued.    LOGARITHMS 


Proportional  Parts. 

Nat 

Nos 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

123 

456 

789 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

1  2  2 

345 

567 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

1  2  2 

345 

567 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

1  2  2 

345 

567 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

1  1  2 

344 

567 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

1  1  2 

344 

567 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

1  1  2 

344 

566 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

1  1  2 

344 

5  6 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

1   2 

334 

5  6 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

1   2 

334 

5  5 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

1   2 

3  3  4 

5  5 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

1   2 

334 

5  5 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248  8254 

1   2 

334 

5  5 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

83128319 

1   2 

334 

556 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

837C 

8376  8382 

1  1  2 

334 

456 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

1  1  2 

234 

456 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

1  1  2 

234 

456 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

1  1  2 

234 

455 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

1  1  2 

234 

455 

73 

8633 

8639 

8645  8651 

8657 

8663 

8669 

8675 

8681 

8686 

1  1  2 

234 

455 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

1  1  2 

234 

4  5 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

1  1  2 

233 

4  5 

76 

8808 

8814 

8820  '8825 

8831 

8837 

8842 

8848 

8854 

8859 

1  1  2 

233 

4  5 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

1  1  2 

233 

4  4 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

1  1  2 

233 

4  4 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

1  1  2 

233 

4  4 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

1  1  2 

233 

445 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

1  1  2 

233 

445 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

1  1  2 

233 

445 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

1  1  2 

233 

445 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

1  1  2 

233 

445 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

1  1  2 

233 

445 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

1  1  2 

233 

445 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

0  1  1 

223 

344 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

0  1  1 

2  2-3 

344 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

0  1  1 

223 

344 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

0  1  1 

223 

344 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

0  1  1 

223 

344 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

0  1  1 

223 

344 

93 

9685 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

0  1  1 

223 

344 

94 

9731 

9?36 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

0  1  1 

223 

344 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

0  1  1 

223 

344 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

0  1  1 

223 

344 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

0  1  1 

223 

344 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

0  1  1 

223 

344 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

0  1  1 

223 

334 

209 


210  APPENDIX 

TABLE  IX 
REFERENCES  ON  ENGINEERING  THERMODYNAMICS 

Engineering   Thermodynamics.     Charles    E.    Lucke.     McGraw-Hill    Book    Co., 

New  York  City,  N.  Y. 
Heat  Engineering.    Arthur  M.  Greene,  Jr.     McGraw-Hill  Book  Co.,  New  York 

City,  N.  Y. 
Applied  Thermodynamics.    Wm.  D.  Ennis.     D.  Van  Nostrand  Co.,  New  York 

City,  N.  Y. 
Principles  of  Thermodynamics.     G.  A.  Goodenough.     H.  Holt  &  Co.,  New  York 

City,  N.  Y. 
Thermodynamics  of  the  Steam  Engine  and  Other  Heat  Engines.    Cecil  H.  Peabody. 

John  Wiley  &  Sons,  Inc.,  New  York  City,  N.  Y. 
Elements  of  Heat  Power  Engineering.    Hirshfeld  and  Barnard.     John  Wiley  & 

Sons,  Inc.,  New  York  City,  N.  Y. 
Thermodynamics.     Dr.  Max  Planck,  Alex.  Ogg,  Trans.     Longmans,  Green  &  Co., 

New  York  City,  N.  Y. 
Thermodynamics.     De  Volson  Wood.     John  Wiley  &  Sons,  Inc.,  New  York  City, 

N.  Y. 

Technical  Thermodynamics.     Gustav  Zeuner,  J.  F.  Klein,  Trans.     D.  Van  Nos- 
trand Co.,  New  York  City,  N.  Y. 
Thermodynamics  of  Heat  Engines.     Sidney  A.  Reeve.     MacMillan  Company, 

New  York  City,  N.  Y. 
An  Introduction  to  General  Thermodynamics.     Henry  A.  Perkins.     John  Wiley  & 

Sons,  Inc.,  New  York  City,  N.  Y. 
Thermodynamics  of  Technical  Gas  Reactions.     Dr.  F.  Haber.     Longmans,  Green 

&  Co.,  New  York  City,  N.  Y. 
The  Thermodynamic  Principles  of  Engine  Design.    L.  M.  Hobbs.     C.  Griffin  & 

Co.,  London. 
An  Outline  of  the  Theory  of  Thermodynamics.     Edgar  Buckingham.     MacMillan 

Company,  New  York  City,  N.  Y. 
The  Temperature-Entropy  Diagram.    Charles  W.  Berry.    John  Wiley  &  Sons,  Inc., 

New  York  City,  N.  Y. 
The  Steam  Engine  and  Other  Steam  Motors.     Robert  C.  H.  Heck.     D.  Van 

Nostrand  Co.,  New  York  City,  N.  Y. 
The  Steam  Engine  and  Turbine.     Robert  C.  H.  Heck.     D.  Van  Nostrand  Co., 

New  York  City,  N.  Y. 
Steam  Engine,  Theory  and  Practice.    Wm.  Ripper.    Longmans,  Green  &  Co., 

New  York  City,  N.  Y. 

Internal  Combustion  Engines.     R.  C.  Carpenter,  and  Diederichs.     D.  Van  Nos- 
trand Co.,  New  York  City,  N.  Y. 
Gas,  Petrol  and  Oil  Engines,  Vol.  I.    Dugald  Clerk.    John  Wiley  &  Sons,  Inc., 

New  York  City,  N.  Y. 
Gas,  Petrol  and  Oil  Engines,  Vol  II.     Clerk  and  Burls.     John  Wiley  &  Sons,  Inc., 

New  York  City,  N.  Y. 
The  Design  and  Construction  of  Internal  Combustion  Engines.    Hugo  Guldner. 

D.  Van  Nostrand  Co.,  New  York  City,  N.  Y. 


APPENDIX  211 

Gas  Engine  Design.    C.  E.  Lucke.    D.  Van  Nostrand  Company,  New  York  City, 

N.  Y. 
Modern  Gas  Engine  and  the  Gas  Producer.     A.  M.  Levin.    John  Wiley  &  Sons, 

Inc.,  New  York  City,  N.  Y. 

The  Gas  Turbine.     H.  Holzwarth.     C.  G.  Griffin  Company,  London. 
Physical  Significance  of  Entropy.     J.  F.  Klein.     D.  Van  Nostrand  Co.,  New  York 

City,  N.  Y. 
Practical  Treatise  on  the  "Otto"  Cycle  Gas  Engine.    Wm.  Norris.    Longmans, 

Green  &  Co.,  New  York  City,  N.  Y. 
Steam  Tables  and  Diagrams.     Marks  &  Davis.    Longmans,  Green  &  Co.,  New 

York  City,  N.  Y. 
Properties  of  Steam  and  Ammonia.     G.  A.  Goodenough,  John  Wiley  &  Sons,  Inc., 

New  York  City,  N.  Y. 
Tables  of  the  Properties  of  Steam  and  Other  Vapors  and  Temperature  Entropy 

Table.     Cecil  H.  Peabody.     John  Wiley  &  Sons,  Inc.,  New  York  City,  N.  Y. 
Thermodynamic  Properties  of  Ammonia.    Robert  B.   Brownlee,   Frederick  G. 

Keyes.     John  Wiley  &  Sons,  Inc.,  New  York  City,  N.  Y. 
Reflections  on  the  Motive  Power  of  Heat  and  on  Machines  fitted  to  Develop  Power. 

N.  L.  S.  Carnot.     Edited  by  R.  H.  Thurston,  John  Wiley  &  Sons,  Inc.,  New 

York  City,  N.  Y. 

Energy.     Sidney  A.  Reeve.    McGraw-Hill  Book  Co.,  New  York  City,  N.  Y. 
A  New  Analysis  of  the  Cylinder  Performance  of  Reciprocating  Engines.     J.  P. 

Clayton,  University  of  Illinois.     Univ.  of  111.  Bulletin,  Vol.  IX,  No.   26. 

Urbana,  111.     1912. 
Thermal  Properties  of  Steam.     G.  A.  Goodenough.     Bulletin  of  the  University  of 

Illinois,  Urbana,  111.    Vol.  XII,  No.  i.     1914. 


INDEX 


Absolute  temperature,  3,  15. 

pressure,  6. 

zero,  3. 

Absorption  system  of  refrigeration,  193. 
Adiabatic  expansion,  27,  32,  36,  43,  94, 
no,  128. 

compression,  32,  36,  43. 

lines  of  steam,  109. 
Air  compressor,  178. 

engines,  49. 

Ammonia  machine,  191. 
Apu,  67. 
Available  energy  of  steam,  128-134. 

Barrel  calorimeter,  84. 
Barrus'  calorimeter,  81,  82. 
Boyle's  law,  15,  29. 
Brake  horse  power,  8. 
Bray  ton  cycle,  57. 
British  thermal  unit,  3. 

Calorie,  3. 

Calorimeter,  steam,  77-85. 

Carnot,  40. 

Carnot  cycle,  40,  117. 

cycle,  efficiency  of,  44. 

cycle,  entropy  changes,  94. 

cycle,  reversed,  45. 
Centigrade,  2,  3. 
Charles'  law,  16. 
Clausius,  119. 

cycle,  119. 

Clayton's  analysis,  147. 
Coefficient  of  flow,  165. 

of  performance,  194. 
Combined  diagrams,  140. 
Combination  law,  16. 
Compound  engine,  140. 
Compressed  air,  178-185. 
Compression,  adiabatic,  32,  36,  43. 


Compression,  isothermal,  27,  42. 

of  gases,  26. 

of  vapors,  104. 
Compressor,  air,  178. 
Condensing  calorimeters,  84. 
Conservation  of  energy,  9. 
Conversion  of  temperatures,  2,  3. 
Cycle,  Brayton,  57. 

Carnot,  40. 

Clausius,  119. 

definition  of,  40. 

hot-air  engine,  49. 

internal-combustion  engine,  53. 

Otto,  56. 

Rankine,  119,  140. 

reversible,  45. 

steam  engine,  125. 

Stirling,  50. 

Density,  65,  154,  198. 

Diagram,  temperature-entropy,  95, 134. 

heat-entropy,  in. 

indicator,  26,  47,  140. 

Mollier,  99,  113,  (Appendix). 

total  heat-entropy,  in,  113. 
Discharge,  maximum,  135. 
Dry  saturated  steam,  71. 
Drying  of  steam  by  throttling,  75. 

Efficiency,  air  engines,  52,  53. 

Carnot  cycle,  44. 

Ericsson  engine,  53. 

Rankine  cycle,  119,  140. 

refrigerating  machine,  94. 

Stirling  engine,  52. 

thermal,  11,  140. 

volumetric,  181,  183. 
Energy,  available,  128-134. 

internal,  21,  35,  69. 

intrinsic,  69. 


213 


214 


INDEX 


Engine,  Ericsson,  52. 

compound,  140. 

hot-air,  49. 

Lenoir,  53. 

Stirling,  50. 
Entropy,  91. 

diagram,  94-102. 

of  the  evaporation,  97. 

of  the  liquid,  96. 
Equivalent  evaporation,  85. 
Ericsson  hot-air  engine,  52. 
Evaporation,  equivalent,  85. 

external  work  of,  67. 

factor  of,  85. 

internal  energy  of,  69. 

latent  heat  of,  67. 
Expansion  of  gases,  26. 

of  vapors,  104. 
Expansions,  adiabatic,  27,  32,  36,  43. 

isothermal,  27,  28,  42,  94. 

poly  tropic,  112. 
External  work,  67.    . 

work  in  steam  formation,  67. 

Factor  of  evaporation,  85. 
First  law  of  thermodynamics,  9. 
Fliegner's  formulas,  155. 
Flow  of  air,  155. 

in  nozzles,  150. 

in  orifices,  150 

of  steam,  167,  171. 
Foot-pound,  6. 
Foot,  square,  6. 

Gas  constant  (R),  18,  22. 
Gas,  perfect,  14,  18. 
Gram-calorie,  3. 
Grashof,  168. 

h  (heat  of  liquid),  66. 

H  (total  heat  of  steam),  68. 

Heat  engine  efficiencies,  40-60. 

of  liquid,  66. 

units,  3. 

Heat-entropy  diagram,  in. 
Hirn's  analysis,  144. 


Hot-air  engine,  49. 

Hyperbolic  logarithms,  31,  (Appendix). 

Horse  power,  8. 

Indicator  diagram,  26,  47,  140. 

diagram,  combined,  140. 
Injectors,  165. 

efficiency  of,  166. 
Internal  energy,  21,  35,  69. 

of  evaporation,  69. 
Intrinsic  energy,  69. 
Isothermal  expansion,  27,  28,  42,  94. 

compression,  27,  42. 

lines  of  steam,  108. 

Joule's  law,  21. 
Kelvin,  21. 

L  (latent  heat  of  steam),  67. 
Latent  heat  of  evaporation,  67. 
Laws  of  perfect  gases,  18. 

of  thermodynamics,  9. 
Logarithms,  natural,  31,  (Appendix). 

"common"  or  ordinary,  31. 
Losses,  turbine,  163. 

Mean  specific  heat,  74,  75. 
Mechanical  equivalent  of  heat,  8. 
Moisture  in  steam,  72,  77. 
Mollier  diagram,  99,  113  (Appendix). 

Naperian  logarithms,  31,  (Appendix). 
Napier's  formula,  168. 
Natural  logarithms,  31,  (Appendix). 
Non-expansive  cycle,  127. 
Non-reversible  cycle,  45. 
Nozzle,  flow  through,  150. 
'  Nozzle  losses,  173. 
Nozzles,  impulse,  162. 
reaction,  164. 

Orifice,  flow  through,  150. 
Otto  cycle,  56. 
Per  cent  wet,  72. 
Perfect  gas,  14,  18. 


INDEX 


215 


Porous  plug  experiment,  21. 
Power  plant  diagrams,  100. 
Pressure- temperature  relation,  16,  36. 

units,  6. 
Properties  of  gases,  197. 

of  steam,  63. 

of  vapors,  62. 

g,  66.. 

Quality  of  steam,  72,  no,  113. 

R  (thermodynamic  or  "gas"  constant), 

18,  22,  48. 
r,67. 

Rankine  cycle,  119,  140. 
Ratio  of  expansion  (r),  30. 
of  specific  heats,  23,  154. 
Receiver  method,  158. 
References,  198. 
Refrigerating  machines,  40,  178. 

media,  87. 
Refrigeration,  applications  of,  178. 

unit  of,  187. 
Regenerator,  51. 
Reversibility,  45. 
Reversible  cycle,  45. 

Saturated  steam,  62,  65. 
Saturation  curve,  142. 
Scales,  thermo metric,  2,  3. 
Second  law  of  thermodynamics,  9. 
Separating  calorimeter,  82. 
Small  calorie,  3. 
Specific  heat,  4,  23. 

heat  at  constant  pressure,  5,  197. 

heat  at  constant  volume,  5,  197. 

heat,  instantaneous  values  of,  76. 

heat,  mean  value  of,  75. 

heat,  ratio  of,  23. 

heat  of  gases,  197. 

heat  of  superheated  steam,  75,  76. 

heat  of  water,  4. 

volume  of  gases,  6,  198. 

volume  of  saturated  steam,  65. 

volume  of  superheated  steam,  75. 
Steam,  dry,  71.  . 


Steam  engines,  efficiencies  of,  119. 

engines,  power  plant,  zoo. 

entropy  for,  96. 

saturated,  71. 

superheated,  72,  73. 

tables,  (Appendix). 

total  heat  of,  68. 

turbine,  163. 

wet,  71. 

Stirling  engine,  50. 
Superheated  steam,  72. 
Superheating  calorimeter,  77,  132. 
Symbols,  be,  x. 

Tables,  steam,  (Appendix). 

vapor,  63. 
Temperature,  absolute,  3. 

-entropy  diagrams,  95-102,  134. 

-entropy  diagram  for  power  plant, 

IOO. 

Thermal  efficiency,  n,  140. 

unit,  British,  3. 

Thermodynamics,  definition  of,  i. 
Thermometric  scales,  2,  3. 
Throttling,  75. 

Throttling  calorimeter,  77,  80,  81. 
Total  heat-entropy  diagram,  in,  113. 

heat  of  saturated  steam,  68. 

heat  of  superheated  steam,  73. 

internal  energy  of  steam,  69. 
Turbine,  steam,  163. 

Under-expansion,  173. 
Unit  of  heat,  3. 

Vaporization  (see  evaporation). 
Vapors,  62,  104. 

properties  of,  62. 
Velocity,  155. 

Volume,  specific,  6,  65,  75,  198. 
Volumetric  efficiency,  181,  183. 

Water,  specific  heat  of,  4. 
Watt,  8. 

Weight,  units  of,  3  (footnote). 
Wet  steam,  71,  no,  131. 


2l6  INDEX 

Wetness  of  steam,  72.  Work,  of  compression,  179. 

Wire-drawing.  75.  of  Rankine  cycle,  120. 
Work,  external,  6,  67. 

of  adiabatic  expansion,  67.  x  (quality  of  steam),  72,  no. 

of  Carnot  cycle,  44. 

of  Clausius  cycle,  120.  Zero,  absolute,  3. 


Velocity 
Ft.  11*.  B.T.U.  Ft.  per 
of  Work  second 

J) o  0  1350 


1.42 


1.46 


1.54  1.58  1.62 

MOLT.IF.R  CHART      RenroHuced  hv 


TOTAL  HEAT-ENTROPY  DIAGRAM. 

The  ordinates  are  Total  Heats. 

The  abscissae  are  Entropies. 

Vertical  lines  are  lines  of  constant  entropy. 

Horizontal  lines  are  lines  of  constant  total  heat. 

Reproduced  by  permission  from 
MASKS  AND  DAVIS*  STZAM  TABLES. 


rlbYP< 


1.74 
nfsion  from  Gebhardt's  "Steam 


1.78 


333C8 


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